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Page No 96:

Question 1:

Answer:

The amount of heat crossing through any cross-section of a slab in time t is called heat current.
It is written as Qt and not as complete derivative dQdt. This is because the amount of heat crossing through any cross section is a function of many variables like temperature difference, area of cross-section, etc. So, we cannot write it as a complete derivative with respect to time.

Page No 96:

Question 2:

The amount of heat crossing through any cross-section of a slab in time t is called heat current.
It is written as Qt and not as complete derivative dQdt. This is because the amount of heat crossing through any cross section is a function of many variables like temperature difference, area of cross-section, etc. So, we cannot write it as a complete derivative with respect to time.

Answer:

Yes, the body will radiate. However, its temperature will not fall down with time because as the temperature of the surroundings is greater than the temperature of the body so, its rate of absorption will be greater than its rate of emission.



Page No 97:

Question 3:

Yes, the body will radiate. However, its temperature will not fall down with time because as the temperature of the surroundings is greater than the temperature of the body so, its rate of absorption will be greater than its rate of emission.

Answer:

Here, major role is played by convection. When we blow air over a spoonful of hot tea, the air coming from our mouth has less temperature than the air above the tea. Since hot air has less density, it rises up and cool air goes down. In this way, the tea cools down.
We know that any hot body radiates. So, the spoonful of tea will also radiate and as the temperature of the surrounding is less then the tea, the tea will cool down with time. Evaporation is also involved in this. On blowing over the hot tea, rate of evaporation increases and the cools down.

Page No 97:

Question 4:

Here, major role is played by convection. When we blow air over a spoonful of hot tea, the air coming from our mouth has less temperature than the air above the tea. Since hot air has less density, it rises up and cool air goes down. In this way, the tea cools down.
We know that any hot body radiates. So, the spoonful of tea will also radiate and as the temperature of the surrounding is less then the tea, the tea will cool down with time. Evaporation is also involved in this. On blowing over the hot tea, rate of evaporation increases and the cools down.

Answer:

No. When the door of the refrigerator is left open in a closed room, the heat given out by the refrigerator to the room will be more than that taken from the room. Therefore, instead of decreasing, the temperature of the room will increase at a slower rate.

Page No 97:

Question 5:

No. When the door of the refrigerator is left open in a closed room, the heat given out by the refrigerator to the room will be more than that taken from the room. Therefore, instead of decreasing, the temperature of the room will increase at a slower rate.

Answer:

We will prefer to seat on a wooden chair because as the conductivity of wood is poorer than that of metal, heat flow from our body to the chair will be less in case of a wooden chair.

Page No 97:

Question 6:

We will prefer to seat on a wooden chair because as the conductivity of wood is poorer than that of metal, heat flow from our body to the chair will be less in case of a wooden chair.

Answer:

Yes, the temperature of the balls can be equalised by radiation. This is because both the spheres will emit radiations in all the directions at different rates.
The ball kept at the temperature of 300 K will gain some thermal energy by the radiation emitted by the ball kept at the temperature of 600 K. Also, it losses energy by radiation.
Similarly, the ball kept at the temperature of 600 K will gain some thermal energy by the radiation emitted by the ball kept at the temperature of 600 K. Also, it losses energy by radiation.
A time comes when the temperature of both the bodies becomes equal.

Yes, the rate of heat gained by the colder sphere is proportional to T24 - T14.

Page No 97:

Question 7:

Yes, the temperature of the balls can be equalised by radiation. This is because both the spheres will emit radiations in all the directions at different rates.
The ball kept at the temperature of 300 K will gain some thermal energy by the radiation emitted by the ball kept at the temperature of 600 K. Also, it losses energy by radiation.
Similarly, the ball kept at the temperature of 600 K will gain some thermal energy by the radiation emitted by the ball kept at the temperature of 600 K. Also, it losses energy by radiation.
A time comes when the temperature of both the bodies becomes equal.

Yes, the rate of heat gained by the colder sphere is proportional to T24 - T14.

Answer:

An ordinary electric fan does not cool the air, still it gives comfort in summer because it circulates the air present in the room. Due to this, evaporation takes place and we feel cooler.

Page No 97:

Question 8:

An ordinary electric fan does not cool the air, still it gives comfort in summer because it circulates the air present in the room. Due to this, evaporation takes place and we feel cooler.

Answer:

The temperature of the atmosphere at a high altitude is around 500°C, but density of air molecule is extremely low at this height. So, very less molecules of air collide with the body of the animal and transfer very less amount of heat. That is why the animal present there would freeze to death instead boiling.

Page No 97:

Question 9:

The temperature of the atmosphere at a high altitude is around 500°C, but density of air molecule is extremely low at this height. So, very less molecules of air collide with the body of the animal and transfer very less amount of heat. That is why the animal present there would freeze to death instead boiling.

Answer:

The heat coming from the sun to us is through the radiation. On colder winter days, when we stand in shade, we do not get the heat of the sun from the radiation. Though we feel cool in the shade, the temperature of the air in shady as well as non-shady regions is the same.

Page No 97:

Question 10:

The heat coming from the sun to us is through the radiation. On colder winter days, when we stand in shade, we do not get the heat of the sun from the radiation. Though we feel cool in the shade, the temperature of the air in shady as well as non-shady regions is the same.

Answer:

During night, the earth's surface radiates infrared radiation of larger wavelength. Gas molecules in the air absorb some of this energy and radiate energy of their own in all directions. Also, water molecules, like the vapour that makes the clouds, absorb more frequencies of infrared energy than clear air does.
Both these factors contribute to the fact that clouds radiate more heat in all directions (including the earth) than clear air does. In turn, this makes the overall temperature on the earth warmer when there is a cloud cover. The heat energy radiated by the earth is reflected back to earth. Due to this, cloudy nights are warmer than the nights with clean sky.

Page No 97:

Question 11:

During night, the earth's surface radiates infrared radiation of larger wavelength. Gas molecules in the air absorb some of this energy and radiate energy of their own in all directions. Also, water molecules, like the vapour that makes the clouds, absorb more frequencies of infrared energy than clear air does.
Both these factors contribute to the fact that clouds radiate more heat in all directions (including the earth) than clear air does. In turn, this makes the overall temperature on the earth warmer when there is a cloud cover. The heat energy radiated by the earth is reflected back to earth. Due to this, cloudy nights are warmer than the nights with clean sky.

Answer:

A white colour dress reflects almost all the radiations falling on it. So, it does not absorb any heat from the sunlight and we feel more comfortable in it. On the other hand, a dark colour dress absorbs maximum radiation falling on it. So, we feel hot in a dark coloured dress during summers.

Page No 97:

Question 1:

A white colour dress reflects almost all the radiations falling on it. So, it does not absorb any heat from the sunlight and we feel more comfortable in it. On the other hand, a dark colour dress absorbs maximum radiation falling on it. So, we feel hot in a dark coloured dress during summers.

Answer:

(d) material of the rod

The thermal conductivity of a rod depends only on the material of the rod. For example, metals are much better conductors than non-metals because metals have large number of free electron that can move freely anywhere in the body of the metal and carry thermal energy from one place to other. Also, 2 copper rods having different lengths and areas of cross-section have same thermal conductivity that depends only on the number of free electrons in copper.

Page No 97:

Question 2:

(d) material of the rod

The thermal conductivity of a rod depends only on the material of the rod. For example, metals are much better conductors than non-metals because metals have large number of free electron that can move freely anywhere in the body of the metal and carry thermal energy from one place to other. Also, 2 copper rods having different lengths and areas of cross-section have same thermal conductivity that depends only on the number of free electrons in copper.

Answer:

(d) by all the three modes

In conduction, heat is transferred from one place to other by vibration of the molecules. In this process, the average position of a molecule does not change. Hence, there is no mass movement of matter.
In convection, heat is transferred from one place to other by actual motion of particles of the medium. When water is heated, hot water moves upwards and cool water moves downwards.
In radiation process, transfer of heat does not require any material medium.
For a room containing air, heat can be transferred via radiation (no medium required) and convection (by the movement of air molecules) and by conduction (due to collision of hot air molecules with other molecules).

Page No 97:

Question 3:

(d) by all the three modes

In conduction, heat is transferred from one place to other by vibration of the molecules. In this process, the average position of a molecule does not change. Hence, there is no mass movement of matter.
In convection, heat is transferred from one place to other by actual motion of particles of the medium. When water is heated, hot water moves upwards and cool water moves downwards.
In radiation process, transfer of heat does not require any material medium.
For a room containing air, heat can be transferred via radiation (no medium required) and convection (by the movement of air molecules) and by conduction (due to collision of hot air molecules with other molecules).

Answer:

(d) T24 - T14

From Stefan-Boltzmann law, the energy of thermal radiation emitted per unit time by a blackbody of surface area A is given by
    u = σAT4
Here, σ is Stefan-Boltzmann constant.
Since the temperature of the solid is less than the surroundings, the temperature of the solid will increase with time until it reaches equilibrium with the surroundings. The rate of emission from the solid will be proportional to T14 and rate of emission from the surroundings will be proportional to T24.
So, the net rate of increase in temperature will be proportional to T24 - T14.

Page No 97:

Question 4:

(d) T24 - T14

From Stefan-Boltzmann law, the energy of thermal radiation emitted per unit time by a blackbody of surface area A is given by
    u = σAT4
Here, σ is Stefan-Boltzmann constant.
Since the temperature of the solid is less than the surroundings, the temperature of the solid will increase with time until it reaches equilibrium with the surroundings. The rate of emission from the solid will be proportional to T14 and rate of emission from the surroundings will be proportional to T24.
So, the net rate of increase in temperature will be proportional to T24 - T14.

Answer:

(b) all bodies

From Stefan-Boltzmann law, the energy of the thermal radiation emitted per unit time by a blackbody of surface area A is given by
     u = σAT4
Here, σ is Stefan-Boltzmann constant.
This law holds true for all the bodies.

Page No 97:

Question 5:

(b) all bodies

From Stefan-Boltzmann law, the energy of the thermal radiation emitted per unit time by a blackbody of surface area A is given by
     u = σAT4
Here, σ is Stefan-Boltzmann constant.
This law holds true for all the bodies.

Answer:

(a) 1 : 1.15

From Stefan-Boltzmann law, energy of the thermal radiation emitted per unit time by a blackbody of surface area A is given by
     u = σAT4
Here, σ is Stefan-Boltzmann constant.

The thermal radiation emitted in a given time by A and B will be in the ratio
uAuB = TA4TB4uAuB=(273+10)4(273+20)4uAuB=11.15

Page No 97:

Question 6:

(a) 1 : 1.15

From Stefan-Boltzmann law, energy of the thermal radiation emitted per unit time by a blackbody of surface area A is given by
     u = σAT4
Here, σ is Stefan-Boltzmann constant.

The thermal radiation emitted in a given time by A and B will be in the ratio
uAuB = TA4TB4uAuB=(273+10)4(273+20)4uAuB=11.15

Answer:

(d) in nonuniform

In steady state, the temperature of the rod is nonuniform maximum at the end near the furnace and minimum at the end that is away from the furnace.

Page No 97:

Question 7:

(d) in nonuniform

In steady state, the temperature of the rod is nonuniform maximum at the end near the furnace and minimum at the end that is away from the furnace.

Answer:

(c) Stefan's law

From Stefan-Boltzman's law, the energy of the thermal radiation emitted per unit time by a blackbody of surface area A is given by,
   u = σAT4
Where σ is Stefan's constant.
Suppose a body at temperature T is kept in a room at temperature T0.

According to Stefan's law, energy of the thermal radiation emitted by the body per unit time is
  u = eσAT4
Here, e is the emissivity of the body.

The energy absorbed per unit time by the body is (due to the radiation emitted by the walls of the room)
 u0 = eσAT04

Thus, the net loss of thermal energy per unit time is
u = u-u0u=eσA(T4-T04)           ...(i)

Newton law of cooling is given by
dTdt = -bA(T-T0)

This can be obtained from equation (i) by considering the temperature difference to be small and doing the binomial expansion.

Page No 97:

Question 8:

(c) Stefan's law

From Stefan-Boltzman's law, the energy of the thermal radiation emitted per unit time by a blackbody of surface area A is given by,
   u = σAT4
Where σ is Stefan's constant.
Suppose a body at temperature T is kept in a room at temperature T0.

According to Stefan's law, energy of the thermal radiation emitted by the body per unit time is
  u = eσAT4
Here, e is the emissivity of the body.

The energy absorbed per unit time by the body is (due to the radiation emitted by the walls of the room)
 u0 = eσAT04

Thus, the net loss of thermal energy per unit time is
u = u-u0u=eσA(T4-T04)           ...(i)

Newton law of cooling is given by
dTdt = -bA(T-T0)

This can be obtained from equation (i) by considering the temperature difference to be small and doing the binomial expansion.

Answer:

(a)
When a hot liquid is kept in a big room, the liquid will loose its temperature with time. The thermal energy emitted by the liquid will be gained by the walls of the room. As the room is big, we can assume that the temperature difference between the room and the liquid is large. From Stephen's law, the liquid emits thermal energy in proportion to T4, where T is the initial temperature of the liquid.
As the temperature decreases, the rate of loss of thermal energy will also decrease. So, the slope of the curve will also decrease.
Therefore, the plot of temperature with time is best represented by the curve in option (a).

Page No 97:

Question 9:

(a)
When a hot liquid is kept in a big room, the liquid will loose its temperature with time. The thermal energy emitted by the liquid will be gained by the walls of the room. As the room is big, we can assume that the temperature difference between the room and the liquid is large. From Stephen's law, the liquid emits thermal energy in proportion to T4, where T is the initial temperature of the liquid.
As the temperature decreases, the rate of loss of thermal energy will also decrease. So, the slope of the curve will also decrease.
Therefore, the plot of temperature with time is best represented by the curve in option (a).

Answer:

(a) a straight line

When a hot liquid is kept in a big room, then the liquid will loose temperature with time. The thermal energy emitted by the liquid will be gained by the walls of the room. As the room is big, we can assume that the temperature difference between the room and the liquid is large. From Stephen's law, the liquid emits thermal energy in proportion to T4, where T is the initial temperature of the liquid. As the temperature decreases, the rate of loss will also decrease. So, the slope of the curve will also decrease. Finally, at equilibrium, the temperature of the room will become equal to the new temperature of the liquid. So, in steady state, the difference between the temperatures of the two will become zero.

A graph is plotted between the logarithm of the numerical value of the temperature difference between the liquid and the room is plotted against time.The logarithm converts the fourth power dependence into a linear dependence with some coefficient (property of log). So, the plot satisfying all the above properties will be a straight line.

Page No 97:

Question 10:

(a) a straight line

When a hot liquid is kept in a big room, then the liquid will loose temperature with time. The thermal energy emitted by the liquid will be gained by the walls of the room. As the room is big, we can assume that the temperature difference between the room and the liquid is large. From Stephen's law, the liquid emits thermal energy in proportion to T4, where T is the initial temperature of the liquid. As the temperature decreases, the rate of loss will also decrease. So, the slope of the curve will also decrease. Finally, at equilibrium, the temperature of the room will become equal to the new temperature of the liquid. So, in steady state, the difference between the temperatures of the two will become zero.

A graph is plotted between the logarithm of the numerical value of the temperature difference between the liquid and the room is plotted against time.The logarithm converts the fourth power dependence into a linear dependence with some coefficient (property of log). So, the plot satisfying all the above properties will be a straight line.

Answer:

(c) more than 5 minutes

Let the temperature of the surrounding be T°C.
Average temperature of the liquid in first case = 62.5°C
Average temperature difference from the surroundings = (62.5 -T)°C

From newton law of cooling,

1°C min -1=-bA(62.5-T)°C-bA = 162.5-Tmin-1         ... (i)

For the second case,
Average temperature = 57.5°C
Temperature difference from the surroundings = (57.5 -T)°C

From Newton's law of cooling and equation (i), 
5°Ct=-bA(57.5-T)°C5°Ct= 162.5-T(57.5-T)°Ct =5( 62.5-T)(57.5-T)t>5 minutes

Page No 97:

Question 1:

(c) more than 5 minutes

Let the temperature of the surrounding be T°C.
Average temperature of the liquid in first case = 62.5°C
Average temperature difference from the surroundings = (62.5 -T)°C

From newton law of cooling,

1°C min -1=-bA(62.5-T)°C-bA = 162.5-Tmin-1         ... (i)

For the second case,
Average temperature = 57.5°C
Temperature difference from the surroundings = (57.5 -T)°C

From Newton's law of cooling and equation (i), 
5°Ct=-bA(57.5-T)°C5°Ct= 162.5-T(57.5-T)°Ct =5( 62.5-T)(57.5-T)t>5 minutes

Answer:

(d) The state of the rod does not change after steady state is reached

The heat transfer will take place from the hot end to the cold end of the rod via conduction. So, with time, the temperature of the rod will increase from the end dipped in boiling water to the end dipped in melting, until it comes in equilibrium with its surroundings. In steady state, the temperature of the rod is nonuniform and constant, maximum at the end dipped in boiling water and minimum at the end dipped in melting ice.
Equilibrium means that the system is stable. So, all the macroscopic variables describing the system will not change with time. Hence, the temperature of the rod will become constant once equilibrium is reached, but its value is different at different positions of the rod.

Page No 97:

Question 2:

(d) The state of the rod does not change after steady state is reached

The heat transfer will take place from the hot end to the cold end of the rod via conduction. So, with time, the temperature of the rod will increase from the end dipped in boiling water to the end dipped in melting, until it comes in equilibrium with its surroundings. In steady state, the temperature of the rod is nonuniform and constant, maximum at the end dipped in boiling water and minimum at the end dipped in melting ice.
Equilibrium means that the system is stable. So, all the macroscopic variables describing the system will not change with time. Hence, the temperature of the rod will become constant once equilibrium is reached, but its value is different at different positions of the rod.

Answer:

(c) reflect radiation
(d) refract radiation

A black body is an ideal concept. A black body is the one that absorbs all the radiation incident on it. So, a black body does not reflect and refract radiation.



Page No 98:

Question 3:

(c) reflect radiation
(d) refract radiation

A black body is an ideal concept. A black body is the one that absorbs all the radiation incident on it. So, a black body does not reflect and refract radiation.

Answer:

(b) convection current

Convection current is the movement of air (or any fluid) due to the difference in the temperatures. During summer days, there is temperature difference of air above the land and river. Due to this, a convection current is set from the river to the land during daytime. On the other hand, during night, a convection current is set from the land to the river. Therefore, a mild air always flows on the shore of a calm river due to the convection current.  

Page No 98:

Question 4:

(b) convection current

Convection current is the movement of air (or any fluid) due to the difference in the temperatures. During summer days, there is temperature difference of air above the land and river. Due to this, a convection current is set from the river to the land during daytime. On the other hand, during night, a convection current is set from the land to the river. Therefore, a mild air always flows on the shore of a calm river due to the convection current.  

Answer:

(c) If both are picked up by bare hands, the steel will be felt hotter than the charcoal
(d) If the two are picked up from the lawn and kept in a cold chamber, the charcoal will lose heat at a faster rate than the steel.

In steel, conductivity is higher than charcoal. So, if both are picked up by bare hands, then heat transfer from the body (steel or charcoal) to our hand will be larger in case of steel. Hence, steel will be hotter than the charcoal.
On the other hand, emissivity of charcoal is higher as compared to steel. So, if the two are picked up from the lawn and kept in a cold chamber, charcoal will lose heat at a faster rate than steel.

Page No 98:

Question 5:

(c) If both are picked up by bare hands, the steel will be felt hotter than the charcoal
(d) If the two are picked up from the lawn and kept in a cold chamber, the charcoal will lose heat at a faster rate than the steel.

In steel, conductivity is higher than charcoal. So, if both are picked up by bare hands, then heat transfer from the body (steel or charcoal) to our hand will be larger in case of steel. Hence, steel will be hotter than the charcoal.
On the other hand, emissivity of charcoal is higher as compared to steel. So, if the two are picked up from the lawn and kept in a cold chamber, charcoal will lose heat at a faster rate than steel.

Answer:

(a) the maximum intensity of radiation will be near the frequency 2v0
(c) the total energy emitted will increase by a factor of 16

From Wein's displacement law,
λmT = b (a constant)

or cTνm=b

Here, T is the absolute temperature of the body.

So, as the temperature is doubled to keep the product on the left hand side constant, frequency is also doubled.

From Stefan's law, we know that the rate of energy emission is proportional to T4.
This implies that total energy emitted will increase by a factor of (2)4, which is equal to 16.

Page No 98:

Question 6:

(a) the maximum intensity of radiation will be near the frequency 2v0
(c) the total energy emitted will increase by a factor of 16

From Wein's displacement law,
λmT = b (a constant)

or cTνm=b

Here, T is the absolute temperature of the body.

So, as the temperature is doubled to keep the product on the left hand side constant, frequency is also doubled.

From Stefan's law, we know that the rate of energy emission is proportional to T4.
This implies that total energy emitted will increase by a factor of (2)4, which is equal to 16.

Answer:

(a) Both will emit equal amount of radiation per unit time in the beginning.
(b) Both will absorb equal amount of radiation from the surrounding in the beginning.

Let the temperature of the surroundings be T0.
From the Stefan-Boltzmann law, the energy of thermal radiation emitted per unit time by a blackbody of surface area A is given by
u = σAT4
Here, σ is Stephen's constant.
Also, the energy absorbed per unit time by the body is given by
u0 = eσAT04
As the two spheres have equal radii and temperatures, their rate of absorption and emission will be equal in the beginning.

Page No 98:

Question 1:

(a) Both will emit equal amount of radiation per unit time in the beginning.
(b) Both will absorb equal amount of radiation from the surrounding in the beginning.

Let the temperature of the surroundings be T0.
From the Stefan-Boltzmann law, the energy of thermal radiation emitted per unit time by a blackbody of surface area A is given by
u = σAT4
Here, σ is Stephen's constant.
Also, the energy absorbed per unit time by the body is given by
u0 = eσAT04
As the two spheres have equal radii and temperatures, their rate of absorption and emission will be equal in the beginning.

Answer:

Given:
Thermal conductivity of the material, k = 0.80 W m–1 °C–1
Area of the cross section of the slab, A = 100 cm2 = 10−2 m2
Thickness of the slab, Δx = 1 cm = 10−2 m

Rate of flow of heat=Temperature differenceThermal resistanceΔQΔt=ΔTΔxk·AΔQΔt=90-10 k·AΔxΔQΔt=80×0.8×10-210-2ΔQΔt=64 J/sΔQΔt=64×60=3840 J/min

Page No 98:

Question 2:

Given:
Thermal conductivity of the material, k = 0.80 W m–1 °C–1
Area of the cross section of the slab, A = 100 cm2 = 10−2 m2
Thickness of the slab, Δx = 1 cm = 10−2 m

Rate of flow of heat=Temperature differenceThermal resistanceΔQΔt=ΔTΔxk·AΔQΔt=90-10 k·AΔxΔQΔt=80×0.8×10-210-2ΔQΔt=64 J/sΔQΔt=64×60=3840 J/min

Answer:

Rate of flow of heat=Temperature differenceThermal resistance
Thickness of the container, l = 1 cm = 10-2 m
Thermal conductivity of the styrofoam sheet, k=0.025 J s-1 m-1 °C-1
Area, A = 0.80 m2
Thermal resistance, lk A=10-20.025×0.80
Temperature difference, T=T1-T2=300-80=220 K

Rate of flow of heat,  ΔQΔt=T1-T2lkA
ΔQΔt=22010-2×0.025×0.80ΔQΔt=440 J/s

Page No 98:

Question 3:

Rate of flow of heat=Temperature differenceThermal resistance
Thickness of the container, l = 1 cm = 10-2 m
Thermal conductivity of the styrofoam sheet, k=0.025 J s-1 m-1 °C-1
Area, A = 0.80 m2
Thermal resistance, lk A=10-20.025×0.80
Temperature difference, T=T1-T2=300-80=220 K

Rate of flow of heat,  ΔQΔt=T1-T2lkA
ΔQΔt=22010-2×0.025×0.80ΔQΔt=440 J/s

Answer:

Rate of flow of heat=Temperature differenceThermal resistance
Temperature of the body, T1=97°F=36.11°C
Temperature of the surroundings, T2=47°F=8.33°C
Conductivity of the cloth, K=0.04 J s-1 m-1 °C-1
Thickness of the cloth, l=0.5 cm=0.005 m
Area of the cloth, A=1.6 m2

Difference in the temperature, T=T1-T2=36.11-8.33=27.78°C
Thermal resistance = lKA=0.0050.04×1.6 = 0.078125

Rate at which heat is flowing out is given by

ΔQΔt=T1-T2lKAΔQΔt=27.780.078125ΔQΔt=356 J/s
 

Page No 98:

Question 4:

Rate of flow of heat=Temperature differenceThermal resistance
Temperature of the body, T1=97°F=36.11°C
Temperature of the surroundings, T2=47°F=8.33°C
Conductivity of the cloth, K=0.04 J s-1 m-1 °C-1
Thickness of the cloth, l=0.5 cm=0.005 m
Area of the cloth, A=1.6 m2

Difference in the temperature, T=T1-T2=36.11-8.33=27.78°C
Thermal resistance = lKA=0.0050.04×1.6 = 0.078125

Rate at which heat is flowing out is given by

ΔQΔt=T1-T2lKAΔQΔt=27.780.078125ΔQΔt=356 J/s
 

Answer:

Area of the bottom of the container, A=25 cm2=25×10-4 m2 
Thickness of the bottom of the container, l = 1 mm = 10-3 m
Latent heat of vaporisation of water, L=2.26×106 J kg-1
Thermal conductivity of the container, k=50 W m-1°C-1
mass = 100 g = 0.1 kg

Rate of heat transfer from the base of the container is given by
ΔQΔt=m LΔt=0.1×2.26×1061 minΔQΔt=0.376×104 J/s   
Also,
ΔQΔt=ΔTlkA0.376×104=T-10010-350×25×10 -40.376×104=50×25×10-4 T-10010-3T-100=3.008×10T=130°C

Page No 98:

Question 5:

Area of the bottom of the container, A=25 cm2=25×10-4 m2 
Thickness of the bottom of the container, l = 1 mm = 10-3 m
Latent heat of vaporisation of water, L=2.26×106 J kg-1
Thermal conductivity of the container, k=50 W m-1°C-1
mass = 100 g = 0.1 kg

Rate of heat transfer from the base of the container is given by
ΔQΔt=m LΔt=0.1×2.26×1061 minΔQΔt=0.376×104 J/s   
Also,
ΔQΔt=ΔTlkA0.376×104=T-10010-350×25×10 -40.376×104=50×25×10-4 T-10010-3T-100=3.008×10T=130°C

Answer:


Given:
Thermal conductivity of the rod, K = 46 J s–1 m–1 °C–1
Length of the rod, l = 1 m
Area of the cross-section of the rod, A = 0.04 cm2
                                                       = 0.04 × 10−4 m2
                                                        = 4 × 10−6 m2
Rate of transfer of heat is given by
ΔQΔt=ΔTlkAΔQΔt=T1-T2 kAlΔQΔt=100-01×46×4×106ΔQΔt=184×10-4 J/s


Also, ΔQ=mLf mtLf=184×10-4mt×3.36×105=184×104m=184×10-43.36×105×tm=5.5×10×10-9×1 kg/sm=5.5×10-8×103 g/sm=5.5×10-5 g/s

Page No 98:

Question 6:


Given:
Thermal conductivity of the rod, K = 46 J s–1 m–1 °C–1
Length of the rod, l = 1 m
Area of the cross-section of the rod, A = 0.04 cm2
                                                       = 0.04 × 10−4 m2
                                                        = 4 × 10−6 m2
Rate of transfer of heat is given by
ΔQΔt=ΔTlkAΔQΔt=T1-T2 kAlΔQΔt=100-01×46×4×106ΔQΔt=184×10-4 J/s


Also, ΔQ=mLf mtLf=184×10-4mt×3.36×105=184×104m=184×10-43.36×105×tm=5.5×10×10-9×1 kg/sm=5.5×10-8×103 g/sm=5.5×10-5 g/s

Answer:

Area of the walls of the box, A = 2400 cm2 = 2400 × 10−4 m2
Thickness of the ice box, l = 2 mm = 2 × 10−3 m
Thermal conductivity of the material of the box, K = 0.06 W m−1 °C−1
Temperature of the water outside the box, T1 = 20°C
Temperature of ice, T2= 0°C

Rate of flow of heat =Temperature differenceThermal resistanceΔQΔt=T1-T2lkAΔQΔt=202×10-3×0.06×2400×10-4ΔQΔt=24×6=144 J/s
Rate at which the ice melts =mLft
ΔQΔt=mtLf144=mt×3.4×105mt=1443.4×105 kg/smt=144×60×603.4×105 kg/hmt=1.52 kg/h

Page No 98:

Question 7:

Area of the walls of the box, A = 2400 cm2 = 2400 × 10−4 m2
Thickness of the ice box, l = 2 mm = 2 × 10−3 m
Thermal conductivity of the material of the box, K = 0.06 W m−1 °C−1
Temperature of the water outside the box, T1 = 20°C
Temperature of ice, T2= 0°C

Rate of flow of heat =Temperature differenceThermal resistanceΔQΔt=T1-T2lkAΔQΔt=202×10-3×0.06×2400×10-4ΔQΔt=24×6=144 J/s
Rate at which the ice melts =mLft
ΔQΔt=mtLf144=mt×3.4×105mt=1443.4×105 kg/smt=144×60×603.4×105 kg/hmt=1.52 kg/h

Answer:

Thickness of porous walls, l = 1 mm = 10−3 m
Mass, m = 10 kg
Latent heat of vapourisation, Lv = 2.27 × 106 J/kg
Thermal conductivity, K = 0.80 J/m s °C
ΔQ = 2.27 × 106 × 10 J
0.1 g of water evaporate in 1 sec, so 10 kg water will evaporate in 105 s.

ΔQΔt=2.27×10 7105ΔQΔt=2.27×102 J/sΔQΔt=ΔTlkAΔQΔt= 42-T10-3·0.80×2×10-22.27×102=42-T10-3×0.80×2×10-2T= 27.8°CT=28°C

Page No 98:

Question 8:

Thickness of porous walls, l = 1 mm = 10−3 m
Mass, m = 10 kg
Latent heat of vapourisation, Lv = 2.27 × 106 J/kg
Thermal conductivity, K = 0.80 J/m s °C
ΔQ = 2.27 × 106 × 10 J
0.1 g of water evaporate in 1 sec, so 10 kg water will evaporate in 105 s.

ΔQΔt=2.27×10 7105ΔQΔt=2.27×102 J/sΔQΔt=ΔTlkAΔQΔt= 42-T10-3·0.80×2×10-22.27×102=42-T10-3×0.80×2×10-2T= 27.8°CT=28°C

Answer:

Thermal conductivity, K = 45 W m–1 °C–1
Length, l = 60 cm = 0.6 m
Area of cross section, A = 0.2 cm2 = 0.2 × 10−4 m2
Initial temperature, T1 = 40°C
Final temperature, T2 = 20°C
Rate of flow of heat=Temerature differenceThermal resistanceΔQΔt=kA(T1-T2)lΔQΔt=45×0.2×10-4 40 - 200.6        =0.03 W

Page No 98:

Question 9:

Thermal conductivity, K = 45 W m–1 °C–1
Length, l = 60 cm = 0.6 m
Area of cross section, A = 0.2 cm2 = 0.2 × 10−4 m2
Initial temperature, T1 = 40°C
Final temperature, T2 = 20°C
Rate of flow of heat=Temerature differenceThermal resistanceΔQΔt=kA(T1-T2)lΔQΔt=45×0.2×10-4 40 - 200.6        =0.03 W

Answer:

Area of cross section, A = 10 cm2 = 10 × 10–4 m2 
Thermal conductivity, K = 200 Js–1 m–1 °C–1
Height, H = 10 cm
Length, l = 1 mm =10–3 m

Temperature inside the cylindrical vessel, T1 = 50°C
Temperature outside the vessel, T2 = 30°C



Rate of flow of heat from 1 flat surface will be given by

ΔQΔt=T1-T2lkAΔQΔt= 50-30×200×10-310-3ΔQΔt=6000 J/sec

Heat escapes out from both the flat surfaces.
Net rate of heat flow = 2 × 6000 = 12000 J/sec

ΔQΔt=m·s·ΔTΔt

Mass = Volume density
10-3×0.1×1000=0.1 g

Using this in the above formula for finding the rate of flow of heat, we get

 12000=0.1×4200×ΔTtΔTΔt= 12000420=28.57As ΔT=1°C1t=28.57t=128.57= 0.035 sec

Page No 98:

Question 10:

Area of cross section, A = 10 cm2 = 10 × 10–4 m2 
Thermal conductivity, K = 200 Js–1 m–1 °C–1
Height, H = 10 cm
Length, l = 1 mm =10–3 m

Temperature inside the cylindrical vessel, T1 = 50°C
Temperature outside the vessel, T2 = 30°C



Rate of flow of heat from 1 flat surface will be given by

ΔQΔt=T1-T2lkAΔQΔt= 50-30×200×10-310-3ΔQΔt=6000 J/sec

Heat escapes out from both the flat surfaces.
Net rate of heat flow = 2 × 6000 = 12000 J/sec

ΔQΔt=m·s·ΔTΔt

Mass = Volume density
10-3×0.1×1000=0.1 g

Using this in the above formula for finding the rate of flow of heat, we get

 12000=0.1×4200×ΔTtΔTΔt= 12000420=28.57As ΔT=1°C1t=28.57t=128.57= 0.035 sec

Answer:



Area of cross section, A = 0.2 cm2 = 0.2 × 10−4 m2
Thermal conductivity, k = 385 W m–1 °C–1

 Rate of flow of heat=Temperature differenceThermal resistanceΔQΔt = kA(T1-T2)lΔQΔt = 80-200.2×385×0.2×10-4         = 2310×10-3        = 2.31 J/sec

Let te temperature of point C be T, which is at a distance of 11 cm from the left end.
Rate of flow of heat is given by

 ΔQΔt = kAΔTlΔTl = ΔQΔt×1kAT-2011×10-2 = 2.31383×0.2×10-4T = 33+20T = 53°C



Page No 99:

Question 11:



Area of cross section, A = 0.2 cm2 = 0.2 × 10−4 m2
Thermal conductivity, k = 385 W m–1 °C–1

 Rate of flow of heat=Temperature differenceThermal resistanceΔQΔt = kA(T1-T2)lΔQΔt = 80-200.2×385×0.2×10-4         = 2310×10-3        = 2.31 J/sec

Let te temperature of point C be T, which is at a distance of 11 cm from the left end.
Rate of flow of heat is given by

 ΔQΔt = kAΔTlΔTl = ΔQΔt×1kAT-2011×10-2 = 2.31383×0.2×10-4T = 33+20T = 53°C

Answer:



One end of the rod is at a temperature of 25°C. So, if no heat current flows through the rod in steady state, then the other end of the rod should also be at a temperature of 25°C.
Let the point at which the other end of the rod is touched be C.
No heat flows through the rod when the temperature at point C is also 25°C.

Heat current through AC = Heat current through CB

 ΔTACxkA = ΔTCB100-xkA100-25x = 25-0100-x3x = 1100-x300-3x = x300 = 4xx = 75 cm

Thus, it should be touched at 75 cm from 100°C end.

Page No 99:

Question 12:



One end of the rod is at a temperature of 25°C. So, if no heat current flows through the rod in steady state, then the other end of the rod should also be at a temperature of 25°C.
Let the point at which the other end of the rod is touched be C.
No heat flows through the rod when the temperature at point C is also 25°C.

Heat current through AC = Heat current through CB

 ΔTACxkA = ΔTCB100-xkA100-25x = 25-0100-x3x = 1100-x300-3x = x300 = 4xx = 75 cm

Thus, it should be touched at 75 cm from 100°C end.

Answer:



Volume of the cube, V = a3 = 216 cm3
Edge of the cube, a = 6 cm
Surface area of the cube = 6a2
                                  = 6 (6 × 10−2)2
                                  = 216 × 10−4 m2
Thickness, l = 0.1 cm = 0.1 × 10–2 m
Temperature difference, Δ T = 5°C
The inner surface of the cube is heated by a 100 W heater.
∴ Power, P = 100 W
Power = Energy per unit time
∴ Rate of flow of heat inside the cube, R = 100 J/s
Rate of flow of heat is given by
 ΔQΔt=ΔTlkA100=k×216×10-4×50.1×10-2k=0.9259 W/m°C

Page No 99:

Question 13:



Volume of the cube, V = a3 = 216 cm3
Edge of the cube, a = 6 cm
Surface area of the cube = 6a2
                                  = 6 (6 × 10−2)2
                                  = 216 × 10−4 m2
Thickness, l = 0.1 cm = 0.1 × 10–2 m
Temperature difference, Δ T = 5°C
The inner surface of the cube is heated by a 100 W heater.
∴ Power, P = 100 W
Power = Energy per unit time
∴ Rate of flow of heat inside the cube, R = 100 J/s
Rate of flow of heat is given by
 ΔQΔt=ΔTlkA100=k×216×10-4×50.1×10-2k=0.9259 W/m°C

Answer:



Temperature of water, T1 = 1°C
Temperature if ice bath, T2 = 0°C
Thermal conductivity, K = 0.5 W/m °C
Length through which heat is lost, l = 2 mm = 2 × 10–3 m
Area of cross section, A = 5 × 10−2 m2
Velocity of the block, v = 10 cm/sec = 0.1 m/s
Let the mass of the block be m.
Power = F · v
           = (mg) v          ......(i)
Also,
 Power=ΔQΔt           .......(ii) ΔQΔt = k·A T1-T2l       ........(iii)

From equation (i), (ii) and (iii), we get
mg v=k·A T1-T2lm=0.5×5×10-2 12×10-3×10×0.1m=12.5 kg

Page No 99:

Question 14:



Temperature of water, T1 = 1°C
Temperature if ice bath, T2 = 0°C
Thermal conductivity, K = 0.5 W/m °C
Length through which heat is lost, l = 2 mm = 2 × 10–3 m
Area of cross section, A = 5 × 10−2 m2
Velocity of the block, v = 10 cm/sec = 0.1 m/s
Let the mass of the block be m.
Power = F · v
           = (mg) v          ......(i)
Also,
 Power=ΔQΔt           .......(ii) ΔQΔt = k·A T1-T2l       ........(iii)

From equation (i), (ii) and (iii), we get
mg v=k·A T1-T2lm=0.5×5×10-2 12×10-3×10×0.1m=12.5 kg

Answer:

Thermal conductivity, K = 1.7 W/m°C
Density of water, ρω = 102 kg/m3
Latent heat of fusion of ice, Lice = 3.36 × 105 J/kg
Length, l = 10 × 10−2 m

(a) Rate of flow of heat is given by
ΔQΔt=T1-T2·KAl
lΔt=T1-T2·KAΔQ=K·A T1-T2m·L= KAT1-T2Al·ρω L= 1.7 0-1010×10-2×103×3.36×105=173.36×10-7=5.059×10-7 m/s=5×10-7 m/s

(b) To form a thin ice layer of thickness dx, let the required be dt.
Mass of that thin layer, dm = A dx ρω
Heat absorbed by that thin layer, dQ = Ldm

dQdt = K·A TxLdmdt = KA Tx(Adxρw)Ldt = KATx0tdt = ρwLK ΔT 00.1x dxt = ρwLK ΔT x2200.1t = ρwLK ΔT×0.122t = 103×3.36×105×0.011.7×10×2t = 3.362×17×106 sect = 3.362×17×1063600 hourst = 27.45 hours

Page No 99:

Question 15:

Thermal conductivity, K = 1.7 W/m°C
Density of water, ρω = 102 kg/m3
Latent heat of fusion of ice, Lice = 3.36 × 105 J/kg
Length, l = 10 × 10−2 m

(a) Rate of flow of heat is given by
ΔQΔt=T1-T2·KAl
lΔt=T1-T2·KAΔQ=K·A T1-T2m·L= KAT1-T2Al·ρω L= 1.7 0-1010×10-2×103×3.36×105=173.36×10-7=5.059×10-7 m/s=5×10-7 m/s

(b) To form a thin ice layer of thickness dx, let the required be dt.
Mass of that thin layer, dm = A dx ρω
Heat absorbed by that thin layer, dQ = Ldm

dQdt = K·A TxLdmdt = KA Tx(Adxρw)Ldt = KATx0tdt = ρwLK ΔT 00.1x dxt = ρwLK ΔT x2200.1t = ρwLK ΔT×0.122t = 103×3.36×105×0.011.7×10×2t = 3.362×17×106 sect = 3.362×17×1063600 hourst = 27.45 hours

Answer:

Let the point upto which ice is formed is at a distance of x m from the top of the lake.
Under steady state, the rate of flow of heat from ice to this point should be equal to the rate flow of heat from water to this point.
Temperature of the top layer of ice = −10°C
Temperature of water at the bottom of the lake = 4°C
Temperature at the point upto which ice is formed = 0°C
ΔQΔtice= ΔQΔtwaterKicexA×10 = Kwater1-xA×41.7×10x=0.5×41-x17x=21-x17-17x=2xx=1719=0.89 mx=89 cm

Page No 99:

Question 16:

Let the point upto which ice is formed is at a distance of x m from the top of the lake.
Under steady state, the rate of flow of heat from ice to this point should be equal to the rate flow of heat from water to this point.
Temperature of the top layer of ice = −10°C
Temperature of water at the bottom of the lake = 4°C
Temperature at the point upto which ice is formed = 0°C
ΔQΔtice= ΔQΔtwaterKicexA×10 = Kwater1-xA×41.7×10x=0.5×41-x17x=21-x17-17x=2xx=1719=0.89 mx=89 cm

Answer:

Thermal conductivity of rod AB, KAB = 50 J/m-s-°C
Temperature of junction at A, TA = 40°C
Thermal conductivity of rod BC, KBC = 200 J/m-s-°C
Temperature of junction at B, TB= 80°C
Thermal conductivity of rod BC, KCA = 400 J/m-s-°C
Temperature of junction at C, TC = 80°C
l = 20 cm = 20 × 10−2 m
A = 1 cm2 = 10−4 m2

(a) ΔQABΔt = KABA TB-TAl
      =50×1×10-4 4020×10-2=1 W

(b) ΔQACΔt = KAC·ATC-TAl
      =400×1×10-4×4020×10-2=8 W

(c) ΔQBCΔt = KBCA TB-TCl
       = 0

Page No 99:

Question 17:

Thermal conductivity of rod AB, KAB = 50 J/m-s-°C
Temperature of junction at A, TA = 40°C
Thermal conductivity of rod BC, KBC = 200 J/m-s-°C
Temperature of junction at B, TB= 80°C
Thermal conductivity of rod BC, KCA = 400 J/m-s-°C
Temperature of junction at C, TC = 80°C
l = 20 cm = 20 × 10−2 m
A = 1 cm2 = 10−4 m2

(a) ΔQABΔt = KABA TB-TAl
      =50×1×10-4 4020×10-2=1 W

(b) ΔQACΔt = KAC·ATC-TAl
      =400×1×10-4×4020×10-2=8 W

(c) ΔQBCΔt = KBCA TB-TCl
       = 0

Answer:

 

Let A be the area of cross section and K be the thermal conductivity of the material of the rod.
Let q1 be the rate of flow of heat through a semicircular rod.
Rate of flow of heat is given by

q1 = dQdt=K·A T1-T2π r

Let q2 be the rate of flow of heat through a straight rod.
q2  = dQdt=KA T1-T22 r

Ratio of the rate of flow of heat through the 2 rods = q1q2=2 rπ r=2π

Page No 99:

Question 18:

 

Let A be the area of cross section and K be the thermal conductivity of the material of the rod.
Let q1 be the rate of flow of heat through a semicircular rod.
Rate of flow of heat is given by

q1 = dQdt=K·A T1-T2π r

Let q2 be the rate of flow of heat through a straight rod.
q2  = dQdt=KA T1-T22 r

Ratio of the rate of flow of heat through the 2 rods = q1q2=2 rπ r=2π

Answer:


Let the temperatures at the ends A and B be TA and TB, respectively.
Rate of flow of heat at end A of the rod is given by
dQAdt = KA.ddlTA
Rate of flow of heat at end B of the rod is given by
dQBdt = KA.ddlTB
Heat absorbed by the rod = ms∆T
Here, s is the specific heat of the rod and ∆T is the temperature difference between ends A and B.

Rate of heat absorption by the rod is given by
dQdt = msdTdt msdTdt = KA.dTAdl-KA.dTBdl0.4.dTdt = 200×1×10-4 5-2.5dTdt = 12.5°C/sec

Page No 99:

Question 19:


Let the temperatures at the ends A and B be TA and TB, respectively.
Rate of flow of heat at end A of the rod is given by
dQAdt = KA.ddlTA
Rate of flow of heat at end B of the rod is given by
dQBdt = KA.ddlTB
Heat absorbed by the rod = ms∆T
Here, s is the specific heat of the rod and ∆T is the temperature difference between ends A and B.

Rate of heat absorption by the rod is given by
dQdt = msdTdt msdTdt = KA.dTAdl-KA.dTBdl0.4.dTdt = 200×1×10-4 5-2.5dTdt = 12.5°C/sec

Answer:


Inner radii = r = 1 cm = 10–2 m
Outer radii = R = 1.2 cm = 1.2 × 10–2 m
Length of the tube, l = 50 cm = 0.5 m
Thermal conductivity, k = 0.15 Js–1 m–1 °C–1

Top View


Let us consider a cylindrical shell of length l, radius x and thickness dx.
Rate of flow of heat q=dQdt
dQdt = -KATdx

Here, the negative sign indicates that the rate of heat flow decreases as x increases.
 q=-K2πxl.dTdxrR dxx=-2πKlqT1T2dTln xrR = -2πKlqT2-T1q = 2πklT1-T2lnRrq = 2π×0.15×0.590ln 1.2×10-21×10-2q = 262.9 J/sec

Page No 99:

Question 20:


Inner radii = r = 1 cm = 10–2 m
Outer radii = R = 1.2 cm = 1.2 × 10–2 m
Length of the tube, l = 50 cm = 0.5 m
Thermal conductivity, k = 0.15 Js–1 m–1 °C–1

Top View


Let us consider a cylindrical shell of length l, radius x and thickness dx.
Rate of flow of heat q=dQdt
dQdt = -KATdx

Here, the negative sign indicates that the rate of heat flow decreases as x increases.
 q=-K2πxl.dTdxrR dxx=-2πKlqT1T2dTln xrR = -2πKlqT2-T1q = 2πklT1-T2lnRrq = 2π×0.15×0.590ln 1.2×10-21×10-2q = 262.9 J/sec

Answer:


Let dθdt be the rate of flow of heat.
Consider an annular ring of radius r and thickness dr.
Rate of flow of heat is given by
dt = -K2πrd.dr                    
Rate of flow of heat is constant.
dt = i 
i = -K2πr.ddrr1r2 drr=-2Kdi θ1θ2 dθln rr1r2=-2πKdiθ2-θ1i=2πKd θ1-θ2ln r2r1

Page No 99:

Question 21:


Let dθdt be the rate of flow of heat.
Consider an annular ring of radius r and thickness dr.
Rate of flow of heat is given by
dt = -K2πrd.dr                    
Rate of flow of heat is constant.
dt = i 
i = -K2πr.ddrr1r2 drr=-2Kdi θ1θ2 dθln rr1r2=-2πKdiθ2-θ1i=2πKd θ1-θ2ln r2r1

Answer:

(a)  When the flat ends are maintained at temperatures T1 and T2 (where T2 > T1):

Area of cross section through which heat is flowing, A=πR22-R12

​Rate of flow of heat=dθdt
                         
                          =KA.Tl=KπR22-R12 T2-T1l

(b)

When the inside of the tube is maintained at temperature T1 and the outside is maintained at T2:

Let us consider a cylindrical shell of radius r and thickness dr.
Rate of flow of heat, q=KA.dTdr
q = KA.dTdrq = K2πrldTdrR1R2 drr = 2πKlqT1T2 dTln rR1R2 = 2πKlq T2-T1ln R2R1 = 2πKlqT2-T1q = 2πKl T2-T1ln R2R1

Page No 99:

Question 22:

(a)  When the flat ends are maintained at temperatures T1 and T2 (where T2 > T1):

Area of cross section through which heat is flowing, A=πR22-R12

​Rate of flow of heat=dθdt
                         
                          =KA.Tl=KπR22-R12 T2-T1l

(b)

When the inside of the tube is maintained at temperature T1 and the outside is maintained at T2:

Let us consider a cylindrical shell of radius r and thickness dr.
Rate of flow of heat, q=KA.dTdr
q = KA.dTdrq = K2πrldTdrR1R2 drr = 2πKlqT1T2 dTln rR1R2 = 2πKlq T2-T1ln R2R1 = 2πKlqT2-T1q = 2πKl T2-T1ln R2R1

Answer:


It is equivalent to the series combination of 2 resistors.
RS = R1 +R2
Resistance of a conducting slab, R=LKA
RS = R1 + R2
L1+L2Ks.A=L1K1A+L2K2AL1+L2KS=L1K1+L2K2L1+L2KS=L1K2+L2K1K1×K2KS=L1+L2 K1 K2L1 K2+L2K1

Page No 99:

Question 23:


It is equivalent to the series combination of 2 resistors.
RS = R1 +R2
Resistance of a conducting slab, R=LKA
RS = R1 + R2
L1+L2Ks.A=L1K1A+L2K2AL1+L2KS=L1K1+L2K2L1+L2KS=L1K2+L2K1K1×K2KS=L1+L2 K1 K2L1 K2+L2K1

Answer:


Rods are connected in series, so the rate of flow of heat is same.

q1 = q2

Rate of flow of heat is given by
q = dQdt = Temperature differenceThermal resistance

As q1 = q2,
T-0Rcu = 100-TRSAK1T-0l  =  100-TlK2A390T = 100-T46T = 10.6°C

Page No 99:

Question 24:


Rods are connected in series, so the rate of flow of heat is same.

q1 = q2

Rate of flow of heat is given by
q = dQdt = Temperature differenceThermal resistance

As q1 = q2,
T-0Rcu = 100-TRSAK1T-0l  =  100-TlK2A390T = 100-T46T = 10.6°C

Answer:



q1 and q2 are heat currents. In other words, they are the rates of flow of heat through aluminium and copper rod, respectively.

Applying KVL at the hot junction, we get
q = q1 + q2
Rate of heat flow, q = KATl

As q = q1 + q2,

KPAT1-T2l=K1AT1-T2l+K2AT1-T2lKp=K1+K2=390+200=590 W/m°Cq=KPAT1-T2lq=590×10-4 60-201q=2.36 W



Page No 100:

Question 25:



q1 and q2 are heat currents. In other words, they are the rates of flow of heat through aluminium and copper rod, respectively.

Applying KVL at the hot junction, we get
q = q1 + q2
Rate of heat flow, q = KATl

As q = q1 + q2,

KPAT1-T2l=K1AT1-T2l+K2AT1-T2lKp=K1+K2=390+200=590 W/m°Cq=KPAT1-T2lq=590×10-4 60-201q=2.36 W

Answer:


Area of cross section, A = 0.20 cm2 = 0.2 × 10–4 m2
Thermal conductivity of aluminium, KAl = 200 W/​m â€‹°C
Thermal conductivity of copper, KCu = 400 W/m​°C
Total heat flowing per second = qAl + qCu
                                         =KAl×A×80-40l+KCu×A×80-40l=200×0.2×10-4×400.2+400×0.2×10-4×400.2=8×10-1+16×10-1=24×10-1=2.4 J/sec
Heat drawn in 1 minute = 2.4 × 60 = 144 J

Page No 100:

Question 26:


Area of cross section, A = 0.20 cm2 = 0.2 × 10–4 m2
Thermal conductivity of aluminium, KAl = 200 W/​m â€‹°C
Thermal conductivity of copper, KCu = 400 W/m​°C
Total heat flowing per second = qAl + qCu
                                         =KAl×A×80-40l+KCu×A×80-40l=200×0.2×10-4×400.2+400×0.2×10-4×400.2=8×10-1+16×10-1=24×10-1=2.4 J/sec
Heat drawn in 1 minute = 2.4 × 60 = 144 J

Answer:



RBC=lKA=5KA  RCD=lKA=60KA  RDE=5KA,  RAB=20KA,   REF=20KA Let: R1=RBC+RCD+RDE=70KALet: RBE=60KA=R2

q = q1 + q2              ...(1)

R1 and R2 are in parallel, so total heat across R1 and R2 will be same.

⇒ q1R1 = q2R2
q1×70KA = q2×60KA7q1 = 6q27q16 = q2          ...2
From equation (1) and (2),
q = q1+7q16q = 13q16q = 130 J130 = 13q16q1 = 60 J/sec

Page No 100:

Question 27:



RBC=lKA=5KA  RCD=lKA=60KA  RDE=5KA,  RAB=20KA,   REF=20KA Let: R1=RBC+RCD+RDE=70KALet: RBE=60KA=R2

q = q1 + q2              ...(1)

R1 and R2 are in parallel, so total heat across R1 and R2 will be same.

⇒ q1R1 = q2R2
q1×70KA = q2×60KA7q1 = 6q27q16 = q2          ...2
From equation (1) and (2),
q = q1+7q16q = 13q16q = 130 J130 = 13q16q1 = 60 J/sec

Answer:


Resistance of any branch, R = lKA

Here, K is the thermal conductivity, A is the area of cross section and l is the length of the conductor.
RBC=l780.A=5×10-2780.ARCD=60×10-2780.ARDE=5×10-2780.ARAB=20×10-2390.AREF=20×10-2390.ARBE=60×10-2390×A

R1=RBC+RCD+RDE=70×10-2780×A

RBE=R2=60×10-2390×A
Since R1and R2are in parallel, the amount of heat flowing through them will be same.
q1R1=q2R2q1q2=R2R1=60×10-2×780×A390×A×70×10-2=127q1q2=127

Page No 100:

Question 28:


Resistance of any branch, R = lKA

Here, K is the thermal conductivity, A is the area of cross section and l is the length of the conductor.
RBC=l780.A=5×10-2780.ARCD=60×10-2780.ARDE=5×10-2780.ARAB=20×10-2390.AREF=20×10-2390.ARBE=60×10-2390×A

R1=RBC+RCD+RDE=70×10-2780×A

RBE=R2=60×10-2390×A
Since R1and R2are in parallel, the amount of heat flowing through them will be same.
q1R1=q2R2q1q2=R2R1=60×10-2×780×A390×A×70×10-2=127q1q2=127

Answer:

(a)

Length, l = 2 mm = 0.0002 m

Rate of flow of heat = Tl/KA                                   =KA.Tl                                   =1×2×40-322×10-3                                   = 8000 J/sec

(b)

Resistance of glass, Rg = lKg.A
Resistance of air, RA = lKA.A

From the circuit diagram, we can find that all the resistors are connected in series.
                              
Rs=Rg+RA+Rg=10-322Kg+1KA=10-3221+10.025=10-32×2×0.025+10.025
Rate of flow of heat, q= TRs                                      = T1-T2Rs                                      =40-32×2×0.02540-3×2×0.025+1                                      = 381W

Page No 100:

Question 29:

(a)

Length, l = 2 mm = 0.0002 m

Rate of flow of heat = Tl/KA                                   =KA.Tl                                   =1×2×40-322×10-3                                   = 8000 J/sec

(b)

Resistance of glass, Rg = lKg.A
Resistance of air, RA = lKA.A

From the circuit diagram, we can find that all the resistors are connected in series.
                              
Rs=Rg+RA+Rg=10-322Kg+1KA=10-3221+10.025=10-32×2×0.025+10.025
Rate of flow of heat, q= TRs                                      = T1-T2Rs                                      =40-32×2×0.02540-3×2×0.025+1                                      = 381W

Answer:



As the rods are connected in series, the rate of flow of heat will be same in both the cases.

Case 1:
Rate of flow of heat is given by
dQdt = KATl
Rate of heat flow in rod P will be same as that in rod Q.

 Kp×A×100-70l = KQ×A×70-0l30 KP = 70 KQKQ = 37 KP       ......(i)
              
Case 2:
Again, the rate of flow of heat will be same in rod P and Q.

 KQ×A× 100-Tl = KP×A×T-0l100 KQ-KQT=KPT100 KQ-KQT=7030KQT        [using (i)]100-T=73T100 = 103TT = 30°C
               

Page No 100:

Question 30:



As the rods are connected in series, the rate of flow of heat will be same in both the cases.

Case 1:
Rate of flow of heat is given by
dQdt = KATl
Rate of heat flow in rod P will be same as that in rod Q.

 Kp×A×100-70l = KQ×A×70-0l30 KP = 70 KQKQ = 37 KP       ......(i)
              
Case 2:
Again, the rate of flow of heat will be same in rod P and Q.

 KQ×A× 100-Tl = KP×A×T-0l100 KQ-KQT=KPT100 KQ-KQT=7030KQT        [using (i)]100-T=73T100 = 103TT = 30°C
               

Answer:

For arrangement (a),


Temperature of the hot end ,T1 = 100°C

Temperature of the cold end ,T2 = 0°C
Rs = R1 + R2 + R3
=lKAlA+lKcuA+lKAlA

=lA1200+1400+1200=lA5400=lA×180
dQdt = q = Rate of flow of heat = T1-T2RS=100-0lA×180Given: q=40 W40=100lA×180lA=200Al=1200

For arrangement (b),


  Rnet = RAl+RCu×RAlRCu+RAl=lKAlA+lKCuA×lKAlllKCuA+lKAlA=lA·KAl+lA KAl+KCu=lA1200+1200+400=lA1200+1600=4600.lA

Rate of flow of heat is given by
q=T1-T2Rnet  =100-04 l600 A  =100×6004×1200=75 W

For arrangement (c),

 

 1Rnet = 1RAl+1RCu+1RAl        =KAlAl+KCuAl+KAlAl1Rnet = KAl+KCu+KAl Al1Rnet=200+400+200×1200Rnet = 200800= 14

Rate of heat flow=ΔTRnet = 10014 =400 W

Page No 100:

Question 31:

For arrangement (a),


Temperature of the hot end ,T1 = 100°C

Temperature of the cold end ,T2 = 0°C
Rs = R1 + R2 + R3
=lKAlA+lKcuA+lKAlA

=lA1200+1400+1200=lA5400=lA×180
dQdt = q = Rate of flow of heat = T1-T2RS=100-0lA×180Given: q=40 W40=100lA×180lA=200Al=1200

For arrangement (b),


  Rnet = RAl+RCu×RAlRCu+RAl=lKAlA+lKCuA×lKAlllKCuA+lKAlA=lA·KAl+lA KAl+KCu=lA1200+1200+400=lA1200+1600=4600.lA

Rate of flow of heat is given by
q=T1-T2Rnet  =100-04 l600 A  =100×6004×1200=75 W

For arrangement (c),

 

 1Rnet = 1RAl+1RCu+1RAl        =KAlAl+KCuAl+KAlAl1Rnet = KAl+KCu+KAl Al1Rnet=200+400+200×1200Rnet = 200800= 14

Rate of heat flow=ΔTRnet = 10014 =400 W

Answer:

 
Let the temperature at junction B be T.
Let q1, q2 and q3 be the heat currents, i.e. rate of flow of heat per unit time in AB, BCE and BDF, respectively.
From the diagram, we can see that
q1 = q2 + q3
The rate of flow of heat is givem by
q = KATl
Using this tn the above equation, we get

KA T1-Tl = KA T-T33l2+KA T-T23l2T1-T = 2 T-T33+2 T-T233T1-T = 2T-2T3+2T-2T2T = -3T1+2 T2+T37

Page No 100:

Question 32:

 
Let the temperature at junction B be T.
Let q1, q2 and q3 be the heat currents, i.e. rate of flow of heat per unit time in AB, BCE and BDF, respectively.
From the diagram, we can see that
q1 = q2 + q3
The rate of flow of heat is givem by
q = KATl
Using this tn the above equation, we get

KA T1-Tl = KA T-T33l2+KA T-T23l2T1-T = 2 T-T33+2 T-T233T1-T = 2T-2T3+2T-2T2T = -3T1+2 T2+T37

Answer:


Given:
KA = KC = K0
KB = KD = 2K0
KE = 3K0, KF = 4K0
K9= 5K0
Here, K denotes the thermal conductivity of the respective rods.

In steady state, temperature at the ends of rod F will be same.

Rate of flow of heat through rod A + rod C = Rate of flow of heat through rod B + rod D

KA·A·T-T1l+KC·A T-T1l = KB·A·T2-Tl+KD·A T2-Tl2K0 T-T1 = 2×2 K0 T2-TTemp of rod F = T = T1+2T23
(b) To find the rate of flow of heat from the source (rod G), which maintains a temperature T2 is given by
Rate of flow of heat, q = TThermal resistance
First, we will find the effective thermal resistance of the circuit.
From the diagram, we can see that it forms a balanced Wheatstone bridge.
Also, as the ends of rod F are maintained at the same temperature, no heat current flows through rod F.
Hence, for simplification, we can remove this branch.

From the diagram, we find that RA and RB are connected in series.
∴ RAB = RA + RB
RC and RD are also connected in series.
∴ RCD = RC + RD
Then, RABand RCDare in parallel connection.
RA = lK0ARB = l2K0ARC = lK0ARD = l2K0ARAB = 3l2K0ARCD = 3l2K0AReff = 3l2K0A×3l2K0A3l2K0A+3l2K0A= 3l4K0A
q = TReff=T1-T23l4K0A=4K0A(T1-T2)3l

Page No 100:

Question 33:


Given:
KA = KC = K0
KB = KD = 2K0
KE = 3K0, KF = 4K0
K9= 5K0
Here, K denotes the thermal conductivity of the respective rods.

In steady state, temperature at the ends of rod F will be same.

Rate of flow of heat through rod A + rod C = Rate of flow of heat through rod B + rod D

KA·A·T-T1l+KC·A T-T1l = KB·A·T2-Tl+KD·A T2-Tl2K0 T-T1 = 2×2 K0 T2-TTemp of rod F = T = T1+2T23
(b) To find the rate of flow of heat from the source (rod G), which maintains a temperature T2 is given by
Rate of flow of heat, q = TThermal resistance
First, we will find the effective thermal resistance of the circuit.
From the diagram, we can see that it forms a balanced Wheatstone bridge.
Also, as the ends of rod F are maintained at the same temperature, no heat current flows through rod F.
Hence, for simplification, we can remove this branch.

From the diagram, we find that RA and RB are connected in series.
∴ RAB = RA + RB
RC and RD are also connected in series.
∴ RCD = RC + RD
Then, RABand RCDare in parallel connection.
RA = lK0ARB = l2K0ARC = lK0ARD = l2K0ARAB = 3l2K0ARCD = 3l2K0AReff = 3l2K0A×3l2K0A3l2K0A+3l2K0A= 3l4K0A
q = TReff=T1-T23l4K0A=4K0A(T1-T2)3l

Answer:



We can infer from the diagram that ΔPQR is similar to ΔPST.
So, by the property of similar triangles,
xl = r-r1r2-r1r2-r1 xl = r-r1r = r1+r2-r1 xlLet: r2-r1l = ar = ax+r1    ..........(i)

Thermal resistance is given by
dR = dxK·AdR=dxK·πr2dR=dxK·π ax+r120RdR=10ldxax+r12R=-1a 1ax+r10lR=-1a 1al+r1-1r1R=-1a 1r2-r1l l+r1-1r1R=-1a 1r2-1r1R=-1r2-r1l×r1-r2r1 r2R=lr1r2Rate of flow of heat=Δ QRq = Q2-Q1l r1r2



Page No 101:

Question 34:



We can infer from the diagram that ΔPQR is similar to ΔPST.
So, by the property of similar triangles,
xl = r-r1r2-r1r2-r1 xl = r-r1r = r1+r2-r1 xlLet: r2-r1l = ar = ax+r1    ..........(i)

Thermal resistance is given by
dR = dxK·AdR=dxK·πr2dR=dxK·π ax+r120RdR=10ldxax+r12R=-1a 1ax+r10lR=-1a 1al+r1-1r1R=-1a 1r2-r1l l+r1-1r1R=-1a 1r2-1r1R=-1r2-r1l×r1-r2r1 r2R=lr1r2Rate of flow of heat=Δ QRq = Q2-Q1l r1r2

Answer:

Given:
Length of the rod, l = 20 cm = 0.2 m
Area of cross section of the rod, A = 1.0 cm2= 1.0×10-4 m2
Thermal conductivity of the material of the rod, k = 200 W m-1°C-1
The temperature of one end of the rod is increased uniformly by 60°C within 10 minutes.
This mean that the rate of increase of the temperature of one end is 0.1°C per second.
 6010×60°C/s 

So, total heat flow can be found by adding heat flow every second.
Rate of flow of heat=dQdtQnet=KAdT2-T1×t

For each interval,
 t=1

Qnet=KA d×0.1+KAd×0.2+KAd×0.3+......+KAd×60.0Qnet=KAd0.1+0.2+.....+60.0

Sum of n terms of an AP is given by
Sn=n2 a+an
Qnet=KAd×60020.1+60 Qnet=200×10-420×10-2×6002×60.1Qnet=1800 J approximately

Page No 101:

Question 35:

Given:
Length of the rod, l = 20 cm = 0.2 m
Area of cross section of the rod, A = 1.0 cm2= 1.0×10-4 m2
Thermal conductivity of the material of the rod, k = 200 W m-1°C-1
The temperature of one end of the rod is increased uniformly by 60°C within 10 minutes.
This mean that the rate of increase of the temperature of one end is 0.1°C per second.
 6010×60°C/s 

So, total heat flow can be found by adding heat flow every second.
Rate of flow of heat=dQdtQnet=KAdT2-T1×t

For each interval,
 t=1

Qnet=KA d×0.1+KAd×0.2+KAd×0.3+......+KAd×60.0Qnet=KAd0.1+0.2+.....+60.0

Sum of n terms of an AP is given by
Sn=n2 a+an
Qnet=KAd×60020.1+60 Qnet=200×10-420×10-2×6002×60.1Qnet=1800 J approximately

Answer:


A = 4πr2
Let:
Radius of the inner sphere = r1
Radius of the outer sphere = r2
Consider a shell of radii r and thickness dr.

For this shell,
Rate of flow of heat, q=-K·4πr2·dTdr
Here, the negative sign indicates that the temperature decreases with increasing radius.
r1r2drr2 = -4πKqT1T2dT
-1rr1r2 = -4πKqT2-T11r1-1r2 = 4πK T1-T2qq=4πKT1-T2 r1 r2r2-r1q=4×227×K 50-10 0.2×0.050.2-0.05K=3W/m°C

Page No 101:

Question 36:


A = 4πr2
Let:
Radius of the inner sphere = r1
Radius of the outer sphere = r2
Consider a shell of radii r and thickness dr.

For this shell,
Rate of flow of heat, q=-K·4πr2·dTdr
Here, the negative sign indicates that the temperature decreases with increasing radius.
r1r2drr2 = -4πKqT1T2dT
-1rr1r2 = -4πKqT2-T11r1-1r2 = 4πK T1-T2qq=4πKT1-T2 r1 r2r2-r1q=4×227×K 50-10 0.2×0.050.2-0.05K=3W/m°C

Answer:

Rate of transfer of heat from the rod is given as
Qt=KAT1-T2l

In time t, the temperature difference becomes half.

In time t, the heat transfer from the rod will be given by
Q=KA(T1-T2)lt   ...(i)

Heat loss by water at temperature T1 is equal to the heat gain by water at temperature T2.

Therefore, heat loss by water at temperature T1in time t is given by
Q=ms(T1'-T1)   ...(ii)

From equation (i) and (ii),

ms(T1-T1')=KA(T1-T2)ltT1'=T1-KA(T1-T2)l(ms)t
This gives us the fall in the temperature of water at temperature T1.

Similarly, rise in temperature of water at temperature T2 is given by
T2'=T2+KA(T1-T2)l(ms)t

Change in the temperature is given by

(T1'-T2')=(T1-T2)-2KA(T1-T2)l(ms)t(T1-T2)-(T1'-T2')=2KA(T1-T2)l(ms)tTt=2KA(T1-T2)l(ms)

Here, Tt is the rate of change of temperature difference.

Taking limit t0,

dTdt=2KA(T1-T2)l(ms)dT(T1-T2)=2KAl(ms)dt

On integrating within proper limit, we get

(T1-T2)(T1-T2)2dT(T1-T2)=2KAl(ms)0tdtln(T1-T2)2(T1-T2)=2KAl(ms)tln[2]=2KAl(ms)tt=ln[2]lms2KA

Page No 101:

Question 37:

Rate of transfer of heat from the rod is given as
Qt=KAT1-T2l

In time t, the temperature difference becomes half.

In time t, the heat transfer from the rod will be given by
Q=KA(T1-T2)lt   ...(i)

Heat loss by water at temperature T1 is equal to the heat gain by water at temperature T2.

Therefore, heat loss by water at temperature T1in time t is given by
Q=ms(T1'-T1)   ...(ii)

From equation (i) and (ii),

ms(T1-T1')=KA(T1-T2)ltT1'=T1-KA(T1-T2)l(ms)t
This gives us the fall in the temperature of water at temperature T1.

Similarly, rise in temperature of water at temperature T2 is given by
T2'=T2+KA(T1-T2)l(ms)t

Change in the temperature is given by

(T1'-T2')=(T1-T2)-2KA(T1-T2)l(ms)t(T1-T2)-(T1'-T2')=2KA(T1-T2)l(ms)tTt=2KA(T1-T2)l(ms)

Here, Tt is the rate of change of temperature difference.

Taking limit t0,

dTdt=2KA(T1-T2)l(ms)dT(T1-T2)=2KAl(ms)dt

On integrating within proper limit, we get

(T1-T2)(T1-T2)2dT(T1-T2)=2KAl(ms)0tdtln(T1-T2)2(T1-T2)=2KAl(ms)tln[2]=2KAl(ms)tt=ln[2]lms2KA

Answer:

Rate of transfer of heat from the rod is given by
Qt=KAT2-T1l

Heat transfer from the rod in time t is given by
Q=KA(T2-T1)lt   ...(i)

Heat loss by the body at temperature T2 is equal to the heat gain by the body at temperature T1.

Therefore, heat loss by the body at temperature T2in time t is given by
Q=m2s2(T2'-T2)   ...(ii)

From equation (i) and (ii),

m2s2(T2-T2')=KA(T2-T1)ltT2'=T2-KA(T2-T1)l(m2s2)t
This gives us the fall in the temperature of the body at temperature T2.

Similarly, rise in temperature of water at temperature T1 is given by
T1'=T1+KA(T2-T1)l(m1s1)t

Change in the temperature is given by

(T2'-T1')=(T2-T1)-KA(T2-T1)lm1 s1t+KA(T2-T1)lm2 s2t(T2'-T1')-(T2-T1)=-KA(T2-T1)lm1 s1t+KA(T2-T1)lm2 s2tTt=-KA(T2-T1)l1m1 s1+1m2 s2t1(T2-T1)T=-KAlm1 s1+m2 s2m1 s1m2 s2

On integrating both the sides, we get

limt01(T2-T1)dT=-KAlm1 s1+m2 s2m1 s1m2 s2dtlnT2-T1=-KAlm1 s1+m2 s2m1 s1m2 s2t(T2-T1)=e-λt
Here, λ=KAlm1 s1+m2 s2m1 s1m2 s2

Page No 101:

Question 38:

Rate of transfer of heat from the rod is given by
Qt=KAT2-T1l

Heat transfer from the rod in time t is given by
Q=KA(T2-T1)lt   ...(i)

Heat loss by the body at temperature T2 is equal to the heat gain by the body at temperature T1.

Therefore, heat loss by the body at temperature T2in time t is given by
Q=m2s2(T2'-T2)   ...(ii)

From equation (i) and (ii),

m2s2(T2-T2')=KA(T2-T1)ltT2'=T2-KA(T2-T1)l(m2s2)t
This gives us the fall in the temperature of the body at temperature T2.

Similarly, rise in temperature of water at temperature T1 is given by
T1'=T1+KA(T2-T1)l(m1s1)t

Change in the temperature is given by

(T2'-T1')=(T2-T1)-KA(T2-T1)lm1 s1t+KA(T2-T1)lm2 s2t(T2'-T1')-(T2-T1)=-KA(T2-T1)lm1 s1t+KA(T2-T1)lm2 s2tTt=-KA(T2-T1)l1m1 s1+1m2 s2t1(T2-T1)T=-KAlm1 s1+m2 s2m1 s1m2 s2

On integrating both the sides, we get

limt01(T2-T1)dT=-KAlm1 s1+m2 s2m1 s1m2 s2dtlnT2-T1=-KAlm1 s1+m2 s2m1 s1m2 s2t(T2-T1)=e-λt
Here, λ=KAlm1 s1+m2 s2m1 s1m2 s2

Answer:

In time dt, heat transfer through the bottom of the cylinder is given by
dQdt=KATs-T0x

For a monoatomic gas, pressure remains constant.

dQ=nCpdTnCpdTdt=KATs-T0x

For a monoatomic gas,
Cp=52R
n5RdT2dt=KATs-T0x
5nR2dTdt=KATs-T0xdTTs-T0=-2KAdt5nRx

Integrating both the sides,
Ts-T0T0T=-2KAt5nRxln Ts-TTs-T0=--2KAt5nRxTs-T=Ts-T0e-2KAt5nRxT=Ts-Ts-T0e-2KAt5nRxT-T0=(Ts-T0)-Ts-T0e-2KAt5nRxT-T0=(Ts-T0)[1-e-2KAt5nRx]

From the gas equation,
PaAlnR=T-T0 PaAlnR=Ts-T0 [1-e-2KAt5nRx]l=nRPaATs-T0 [1-e-2KAt5nRx]

Page No 101:

Question 39:

In time dt, heat transfer through the bottom of the cylinder is given by
dQdt=KATs-T0x

For a monoatomic gas, pressure remains constant.

dQ=nCpdTnCpdTdt=KATs-T0x

For a monoatomic gas,
Cp=52R
n5RdT2dt=KATs-T0x
5nR2dTdt=KATs-T0xdTTs-T0=-2KAdt5nRx

Integrating both the sides,
Ts-T0T0T=-2KAt5nRxln Ts-TTs-T0=--2KAt5nRxTs-T=Ts-T0e-2KAt5nRxT=Ts-Ts-T0e-2KAt5nRxT-T0=(Ts-T0)-Ts-T0e-2KAt5nRxT-T0=(Ts-T0)[1-e-2KAt5nRx]

From the gas equation,
PaAlnR=T-T0 PaAlnR=Ts-T0 [1-e-2KAt5nRx]l=nRPaATs-T0 [1-e-2KAt5nRx]

Answer:

Given:
Area of the body, A = 1.6 m2
Temperature of the body, T = 310 K
From Stefan-Boltzmann law,

Energy radiatedTime= σAT4

Here, A is the area of the body and σ is the Stefan-Boltzmann constant.

Energy radiated per second = 1.6 × 6 × 10−8 × (310)4
                                              = 886.58 887 J

Page No 101:

Question 40:

Given:
Area of the body, A = 1.6 m2
Temperature of the body, T = 310 K
From Stefan-Boltzmann law,

Energy radiatedTime= σAT4

Here, A is the area of the body and σ is the Stefan-Boltzmann constant.

Energy radiated per second = 1.6 × 6 × 10−8 × (310)4
                                              = 886.58 887 J

Answer:

Given:
Area of the body, A = 12 × 10−4 m2
Temperature of the body, T = (273 + 20) = 293 K
Emissivity, e = 0.80
Rate of emission of heat, R = AeσT4
R = 12 × 10−4 × 0.80 × 6.0 × 10−8 × (293)4
R = 0.42 J (approximately)

Page No 101:

Question 41:

Given:
Area of the body, A = 12 × 10−4 m2
Temperature of the body, T = (273 + 20) = 293 K
Emissivity, e = 0.80
Rate of emission of heat, R = AeσT4
R = 12 × 10−4 × 0.80 × 6.0 × 10−8 × (293)4
R = 0.42 J (approximately)

Answer:

(a) Rate of loss of heat = eAσT4

Rate of loss of heatAlRate of loss of heatC=eAl σ·T4×4πr2eC σ T4·4π 2r2                                        =1 : 4       (as eAl = eC)

(b)  Relation between the amount of heat loss by both the spheres in a small time t is given by

ΔQ1=4×ΔQ2m1 s1 ΔT1=4m2 s2 ΔT2ΔT1ΔT2=4m2s2m1s1ΔT1ΔT2=4×ρ2×43π(2r)3×s2ρ1×43πr3×s1ΔT1ΔT2=4×3.4ρ1×8×390ρ1×900=47.14:1

Page No 101:

Question 42:

(a) Rate of loss of heat = eAσT4

Rate of loss of heatAlRate of loss of heatC=eAl σ·T4×4πr2eC σ T4·4π 2r2                                        =1 : 4       (as eAl = eC)

(b)  Relation between the amount of heat loss by both the spheres in a small time t is given by

ΔQ1=4×ΔQ2m1 s1 ΔT1=4m2 s2 ΔT2ΔT1ΔT2=4m2s2m1s1ΔT1ΔT2=4×ρ2×43π(2r)3×s2ρ1×43πr3×s1ΔT1ΔT2=4×3.4ρ1×8×390ρ1×900=47.14:1

Answer:

Given:
Power of the bulb, P = 100 W
Length of the filament, l = 1.0 m
Radius of the filament, r = 4 × 10−5 m

According to Stefan's law,

Energy radiatedTime=e·AσT4

Here, e is the emissivity of the tungsten and σ is Stefan's constant.

 P=e·(πr2)σT4
100=0.8×3.14×4×10-52×6×10-8×T4T=1700 K  (approximately)

Page No 101:

Question 43:

Given:
Power of the bulb, P = 100 W
Length of the filament, l = 1.0 m
Radius of the filament, r = 4 × 10−5 m

According to Stefan's law,

Energy radiatedTime=e·AσT4

Here, e is the emissivity of the tungsten and σ is Stefan's constant.

 P=e·(πr2)σT4
100=0.8×3.14×4×10-52×6×10-8×T4T=1700 K  (approximately)

Answer:

(a)
Area of the ball, A = 20 × 10−4 m2
Temperature of the ball, T = 57°C = 57 + 273 = 330 K
Amount of heat radiated per second = AσT4
   = 20 × 10−4 × 6 × 10−8 × (330)4
   = 1.42 J

(b)
Net rate of heat flow from the ball when its temperature is 200°C is given by

eAσ T14-T24=20×10-4×6×10-8×1 4734-3304                 [ e=1]=4.58 W

Page No 101:

Question 44:

(a)
Area of the ball, A = 20 × 10−4 m2
Temperature of the ball, T = 57°C = 57 + 273 = 330 K
Amount of heat radiated per second = AσT4
   = 20 × 10−4 × 6 × 10−8 × (330)4
   = 1.42 J

(b)
Net rate of heat flow from the ball when its temperature is 200°C is given by

eAσ T14-T24=20×10-4×6×10-8×1 4734-3304                 [ e=1]=4.58 W

Answer:

Given:
Radius of the spherical tungsten, r = 10−2 m
Emissivity of the tungsten, e = 0.3
Stefan's constant, σ = 6.0 × 10−8 W m−2 K−4
Surface area of the spherical tungsten, A = 4πr2
A = 4π (10−2)2
A = 4π × 10−4 m2

Rate at which energy must be supplied is given by 

eσAT24-T14
= (0.3) × 6 × 10−8 × 4π × 10−4 × ((1000)4 − (300)4)
= 22.42  watt = 22 watt

Page No 101:

Question 45:

Given:
Radius of the spherical tungsten, r = 10−2 m
Emissivity of the tungsten, e = 0.3
Stefan's constant, σ = 6.0 × 10−8 W m−2 K−4
Surface area of the spherical tungsten, A = 4πr2
A = 4π (10−2)2
A = 4π × 10−4 m2

Rate at which energy must be supplied is given by 

eσAT24-T14
= (0.3) × 6 × 10−8 × 4π × 10−4 × ((1000)4 − (300)4)
= 22.42  watt = 22 watt

Answer:

It is given that a cube behaves like a black body.
∴ Emissivity, e = 1
Stefan's constant, σ = 6 × 10−8 W/(m2 K4)
Surface area of the cube, A = 6 × 25 × 10−4
Mass of the cube, m = 1 kg
Specific heat capacity of the material of the cube, s = 400 J/(kg-K)
Temperature of the cube, T1 = 227 + 273 = 500 K
Temperature of the surrounding, T0 = 27 + 273 = 300 K
Rate of flow of heat is given by
ΔQΔt=eAσT4-T04ms·ΔTΔt=1×6×10-8×6×25×10-4 5004-3004ΔTΔt =36×25×5004-3004×10-12400ΔTΔt =0.12°C/s

Page No 101:

Question 46:

It is given that a cube behaves like a black body.
∴ Emissivity, e = 1
Stefan's constant, σ = 6 × 10−8 W/(m2 K4)
Surface area of the cube, A = 6 × 25 × 10−4
Mass of the cube, m = 1 kg
Specific heat capacity of the material of the cube, s = 400 J/(kg-K)
Temperature of the cube, T1 = 227 + 273 = 500 K
Temperature of the surrounding, T0 = 27 + 273 = 300 K
Rate of flow of heat is given by
ΔQΔt=eAσT4-T04ms·ΔTΔt=1×6×10-8×6×25×10-4 5004-3004ΔTΔt =36×25×5004-3004×10-12400ΔTΔt =0.12°C/s

Answer:

According to Stefan,s law,
Power = eAσT4-T04
Temperature difference = 200 K

Let the emissivity of copper be e.

210 = eAσ(5004 − 3004)   ...(1)

When the surface of the copper sphere is completely blackened, 700 W is needed to maintain the temperature of the sphere.
For a black body,
e = 1
700 = 1·(5004 − 3004)   ...(2)
On dividing equation (1) by equation (2) we have,

210700=e1e=0.3

Page No 101:

Question 47:

According to Stefan,s law,
Power = eAσT4-T04
Temperature difference = 200 K

Let the emissivity of copper be e.

210 = eAσ(5004 − 3004)   ...(1)

When the surface of the copper sphere is completely blackened, 700 W is needed to maintain the temperature of the sphere.
For a black body,
e = 1
700 = 1·(5004 − 3004)   ...(2)
On dividing equation (1) by equation (2) we have,

210700=e1e=0.3

Answer:

Given:
Surface area of the spherical ball, SA = 20 cm2 = 20×10-4 m2
Surface area of the spherical shell, SB = 80 cm2 = 80×10-4 m2
Temperature of the spherical ball, TA = 300 K
Temperature of the spherical shell, TB = 300 K

Radiation energy emitted per second by the spherical ball A is given by
EA=σSATA4EA=6.0×10-8×20×10-4×3004EA=0.97 J

Radiation energy emitted per second by the inner surface of the spherical shell B is given by
EB=σSBTB4EB=6.0×10-8×80×10-4×3004EB=3.76 J 3.8 J

Energy emitted by the inner surface of B that falls back on its surface is given by

E=EB-EA=3.76-0.94E=2.82 J

Page No 101:

Question 48:

Given:
Surface area of the spherical ball, SA = 20 cm2 = 20×10-4 m2
Surface area of the spherical shell, SB = 80 cm2 = 80×10-4 m2
Temperature of the spherical ball, TA = 300 K
Temperature of the spherical shell, TB = 300 K

Radiation energy emitted per second by the spherical ball A is given by
EA=σSATA4EA=6.0×10-8×20×10-4×3004EA=0.97 J

Radiation energy emitted per second by the inner surface of the spherical shell B is given by
EB=σSBTB4EB=6.0×10-8×80×10-4×3004EB=3.76 J 3.8 J

Energy emitted by the inner surface of B that falls back on its surface is given by

E=EB-EA=3.76-0.94E=2.82 J

Answer:

Length, l = 50 cm
Cross sectional area, A =1 cm2
Stefan's constant, σ = 6 × 10−8 W m−2 K−4
Temperature of the blackened end = 17°C
Temperature of the chamber =  27°C
Temperature of one end of the rod = 0°C

According to Stefan's law,

ΔQΔt = ε·A σ T24-T14ΔQΔt·A = ε σ T24-T14ΔQΔt·A = 1×6×10-8 3004-2904=6×10.3                           ...   1 

Also, ΔQΔt=K A T1-T2lΔQΔt·A=K×170.5          ...        2From 1 and 2,6×10.3 = K×170.5K=1.8 W/°C



Page No 102:

Question 49:

Length, l = 50 cm
Cross sectional area, A =1 cm2
Stefan's constant, σ = 6 × 10−8 W m−2 K−4
Temperature of the blackened end = 17°C
Temperature of the chamber =  27°C
Temperature of one end of the rod = 0°C

According to Stefan's law,

ΔQΔt = ε·A σ T24-T14ΔQΔt·A = ε σ T24-T14ΔQΔt·A = 1×6×10-8 3004-2904=6×10.3                           ...   1 

Also, ΔQΔt=K A T1-T2lΔQΔt·A=K×170.5          ...        2From 1 and 2,6×10.3 = K×170.5K=1.8 W/°C

Answer:

Stefan's constant, σ = 6 × 10−8 W m−2 K−4
l = 0.2 m
T1 = 300 K
T2 = 750 K
 ΔQΔt = εAσ T24-T14ΔQΔt A = εσ T24-T14        ...    1Also, ΔQΔt = K A θ1-θ2l           ΔQΔt·A= K 800-7500.2   ...  2From 1 and 2,εσT24-T14=K×500.21×6×10-8 7504-3004=K×500.2K=74 w/m °C

Page No 102:

Question 50:

Stefan's constant, σ = 6 × 10−8 W m−2 K−4
l = 0.2 m
T1 = 300 K
T2 = 750 K
 ΔQΔt = εAσ T24-T14ΔQΔt A = εσ T24-T14        ...    1Also, ΔQΔt = K A θ1-θ2l           ΔQΔt·A= K 800-7500.2   ...  2From 1 and 2,εσT24-T14=K×500.21×6×10-8 7504-3004=K×500.2K=74 w/m °C

Answer:

Given:
Volume of water, V = 100 cc = 100×10-3 m3
Change in the temperature of the liquid, ∆θ = 5°C
Time, T = 5 min

For water,

msθt=KAl(T1-T0)
mst=KAl(T1-T0)θVρst=KAl(T1-T0)θ100×10-3×1000×42005=KAl(313-T0)θ  ...(i)

For K-oil,

msdt=KAl9(T1-T0)θ
mst=KAl(T1-T0)θVρst=KAl(T1-T0)θ100×10-3×800×2100t=KAl(313-T0)θ  ...(ii)

From (i) and (ii),

100×10-3×800×2100t=100×10-3×1000×42005t=5×800×21001000×4200=20001000t=2 min

Page No 102:

Question 51:

Given:
Volume of water, V = 100 cc = 100×10-3 m3
Change in the temperature of the liquid, ∆θ = 5°C
Time, T = 5 min

For water,

msθt=KAl(T1-T0)
mst=KAl(T1-T0)θVρst=KAl(T1-T0)θ100×10-3×1000×42005=KAl(313-T0)θ  ...(i)

For K-oil,

msdt=KAl9(T1-T0)θ
mst=KAl(T1-T0)θVρst=KAl(T1-T0)θ100×10-3×800×2100t=KAl(313-T0)θ  ...(ii)

From (i) and (ii),

100×10-3×800×2100t=100×10-3×1000×42005t=5×800×21001000×4200=20001000t=2 min

Answer:

Let the temperature of the surroundings be T0°C.

Case 1:
Initial temperature of the body = 50°C
Final temperature of the body = 45°C
Average temperature = 47.5 °C

Difference in the temperatures of the body and its surrounding = (47.5 − T)°C
Rate of fall of temperature = Tt = 55  = 1°C/min

By Newton's law of cooling,

dTdt=-K Tavg-T01=-K 47.5-T0       ...  1

Case 2:
Initial temperature of the body = 45°C
Final temperature of the body = 40°C
Average temperature = 42.5 °C

Difference in the temperatures of the body and its surrounding = (42.5 − T0)°C
Rate of fall of temperature = Tt = 58 = 58°C/min
From Newton,s law of cooling,
dTdt=-K Tavg-T00.625=-K 42.5-T0       ...  2

Dividing (1) by (2),

10.625 =47.5-T042.5-T042.5-T0 = 29.68-0.625TT0 = 34°C

Page No 102:

Question 52:

Let the temperature of the surroundings be T0°C.

Case 1:
Initial temperature of the body = 50°C
Final temperature of the body = 45°C
Average temperature = 47.5 °C

Difference in the temperatures of the body and its surrounding = (47.5 − T)°C
Rate of fall of temperature = Tt = 55  = 1°C/min

By Newton's law of cooling,

dTdt=-K Tavg-T01=-K 47.5-T0       ...  1

Case 2:
Initial temperature of the body = 45°C
Final temperature of the body = 40°C
Average temperature = 42.5 °C

Difference in the temperatures of the body and its surrounding = (42.5 − T0)°C
Rate of fall of temperature = Tt = 58 = 58°C/min
From Newton,s law of cooling,
dTdt=-K Tavg-T00.625=-K 42.5-T0       ...  2

Dividing (1) by (2),

10.625 =47.5-T042.5-T042.5-T0 = 29.68-0.625TT0 = 34°C

Answer:

Let water equivalent to calorimeter be w.
Change in temperature = 5°C
Specific heat of water = 4200 J/Kg °C

Rate of flow of heat is given by

q = Energy per unit time = msTt

Case 1:

q1 = w+50×10-3×4200×510

Case 2

 q2 = w+100×10-3×4200×518

From calorimeter theory, these two rates of flow of heat should be equal to each other.
q1 = q2

w+50×10-3×4200×510 = w+100×10-3×4200×518
18 (w + 50 × 10−3) = 10 (w + 100 × 10−3)
w = 12.5 × 10−3 kg
w = 12.5 g

Page No 102:

Question 53:

Let water equivalent to calorimeter be w.
Change in temperature = 5°C
Specific heat of water = 4200 J/Kg °C

Rate of flow of heat is given by

q = Energy per unit time = msTt

Case 1:

q1 = w+50×10-3×4200×510

Case 2

 q2 = w+100×10-3×4200×518

From calorimeter theory, these two rates of flow of heat should be equal to each other.
q1 = q2

w+50×10-3×4200×510 = w+100×10-3×4200×518
18 (w + 50 × 10−3) = 10 (w + 100 × 10−3)
w = 12.5 × 10−3 kg
w = 12.5 g

Answer:

In steady state, the body has reached equilibrium. So, no more heat will be exchanged between the body and the surrounding.

This implies that at steady state,
Rate of loss of heat = Rate at which heat is supplied

Given: 
Mass, m = 1 kg
Power of the heater = 20 W
Room temperature = 20°C

(a)At steady state,
Rate of loss of heat = Rate at which heat is supplied
And, rate of loss/gain of heat = Power

 dQdt = P = 20 W

(b) By Newton's law of cooling, rate of cooling is directly proportional to the difference in temperature.
So, when the body is in steady state, then its rate of cooling is given as

dQdt = K T-T020 = K 50-20 K=23
When the temperature of the body is 30°C, then its rate of cooling is given as
dQdt = K T-T0=2330-20=203W
The initial rate of cooling when the body,s temperature is 20°C is given as

dQdt20=0  dQdt30=203

dQdtavg=103
t = 5 min = 300 s
Heat liberated=103×300=1000 J
Net heat absorbed = Heat supplied − Heat Radiated
                               = 6000 − 1000 = 5000 J
(d) Net heat absorbed is used for raising the temperature of the body by 10°C.
∴ m S ∆T = 5000
S=5000m×T=50001×10=500 J Kg-1C-1

Page No 102:

Question 54:

In steady state, the body has reached equilibrium. So, no more heat will be exchanged between the body and the surrounding.

This implies that at steady state,
Rate of loss of heat = Rate at which heat is supplied

Given: 
Mass, m = 1 kg
Power of the heater = 20 W
Room temperature = 20°C

(a)At steady state,
Rate of loss of heat = Rate at which heat is supplied
And, rate of loss/gain of heat = Power

 dQdt = P = 20 W

(b) By Newton's law of cooling, rate of cooling is directly proportional to the difference in temperature.
So, when the body is in steady state, then its rate of cooling is given as

dQdt = K T-T020 = K 50-20 K=23
When the temperature of the body is 30°C, then its rate of cooling is given as
dQdt = K T-T0=2330-20=203W
The initial rate of cooling when the body,s temperature is 20°C is given as

dQdt20=0  dQdt30=203

dQdtavg=103
t = 5 min = 300 s
Heat liberated=103×300=1000 J
Net heat absorbed = Heat supplied − Heat Radiated
                               = 6000 − 1000 = 5000 J
(d) Net heat absorbed is used for raising the temperature of the body by 10°C.
∴ m S ∆T = 5000
S=5000m×T=50001×10=500 J Kg-1C-1

Answer:

Given:
Heat capacity of the metal block, s = 80 J°C−1
Heat absorb by the metal block is,
H=s×T=80×(30-10)=800 J

Rate of rise in temperature of the block=dθdt = 2°C/sRate of fall in temperature of the block=dθdt = -0.2°C/s
The negative sign indicates that the temperature is falling with time.

(a) Energy = s (θ)
Power = Energy per unit time

∴ Power of the heater=Heat capacity×dθdt
 P = 80 × 2
 P = 160 W

(b) Power Radiated = Energy lost per unit time.
 P' = Heat capacity×dθdt
Here, dθdt represents the rate of decrease in temperature.
    P' = 80 × 0.2
    P' = 16 W

(c) 16 W of power is radiated when the temperature of the block decreases from 30°C to 20°C.
 s×dθdtdec=K θ-θ0
⇒ 16 = K (30 − 20)
     K = 1.6

From newton's law of cooling,
 sdθdt = K θ-θ0sdθdt=1.6 30-25sdθdt=1.6×5sdθdt=8 Watt

Page No 102:

Question 55:

Given:
Heat capacity of the metal block, s = 80 J°C−1
Heat absorb by the metal block is,
H=s×T=80×(30-10)=800 J

Rate of rise in temperature of the block=dθdt = 2°C/sRate of fall in temperature of the block=dθdt = -0.2°C/s
The negative sign indicates that the temperature is falling with time.

(a) Energy = s (θ)
Power = Energy per unit time

∴ Power of the heater=Heat capacity×dθdt
 P = 80 × 2
 P = 160 W

(b) Power Radiated = Energy lost per unit time.
 P' = Heat capacity×dθdt
Here, dθdt represents the rate of decrease in temperature.
    P' = 80 × 0.2
    P' = 16 W

(c) 16 W of power is radiated when the temperature of the block decreases from 30°C to 20°C.
 s×dθdtdec=K θ-θ0
⇒ 16 = K (30 − 20)
     K = 1.6

From newton's law of cooling,
 sdθdt = K θ-θ0sdθdt=1.6 30-25sdθdt=1.6×5sdθdt=8 Watt

Answer:

According to Newton's law of cooling,
dθdt=-kθ-θ0

(a) Maximum heat that the body can lose, ∆Qmax = ms1 − θ0)

(b) If the body loses 90% of the maximum heat, then the fall in temperature will be θ.
Qmax×90100=ms(θ1-θ)ms(θ1-θ0)×910=ms(θ1-θ)θ=θ1-(θ1-θ0)×910θ=θ1-9θ010    ...(i)

From Newton's law of cooling,
 dθdt=-kθ1-θ
Integrating this equation within the proper limit, we get
At time t = 0,
   θ = θ1
At time t,
   θ = θ

θ1θdθθ1-θ=-k0t dtln(θ1-θθ1-θ0)=-kt θ1-θ=θ1-θ0e-kt        ...ii

From (i) and (ii),

θ1-9θ010-θ0=θ1-θ0 e-ktt=ln (10k)



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