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#### Page No 46:

#### Question 1:

Is heat a conserved quantity?

#### Answer:

Yes, heat is a conserved quantity. Suppose two bodies A and B are at different temperatures, wherein A is higher and B is lower. When they are brought in contact, the heat given by A is equal to the heat gained by B. Thus, the heat is conserved in the system.

Here, we have ignored the heat exchanged with the surroundings. If we consider the surroundings, the heat of the system will be conserved.

#### Page No 46:

#### Question 2:

The calorie is defined as 1 cal = 4.186 joule. Why not as 1 cal = 4 J to make the conversions easy?

#### Answer:

One calorie is defined as the heat required to raise the temperature of 1 g of water from 14.5^{o} C to 15.5^{o} C. Also, exactly 4.186 J of work is to be done on heating 1 g of water from 14.5^{o} C to 15.5^{o} C. So, we cannot round-off 4.186 J to 4 J as it will give significant difference in work done required to raise the temperature and heat required to raise the temperature.

Since work and heat are equivalent here, so on taking 1 calorie = 4 J, the validity of the equivalence of work and heat will get defied. Thus, just to make conversion easy, 1 calorie cannot be taken equal to 4 J.

#### Page No 46:

#### Question 3:

A calorimeter is kept in a wooden box to insulate it thermally from the surroundings. Why is it necessary?

#### Answer:

A calorimeter is kept in a wooden box to insulate it thermally from the surroundings because in order to determine the specific heat capacity, the total heat transferred must be known. Heat must not be exchanged with the surroundings, otherwise the principle of calorimeter would not stand valid.

#### Page No 46:

#### Question 4:

In a calorimeter, the heat given by the hot object is assumed to be equal to the heat taken by the cold object. Does it mean that heat of the two objects taken together remains constant?

#### Answer:

Yes, heat of the two objects taken together remains constant. If no heat is lost to the surroundings, the heat of the two bodies taken together actually remains conserved.

#### Page No 46:

#### Question 5:

In Regnault's apparatus for measuring specific heat capacity of a solid, there is an inlet and an outlet in the steam chamber. The inlet is near the top and the outlet is near the bottom. Why is it better than the opposite choice where the inlet is near the bottom and the outlet is near the top?

#### Answer:

The inlet is near the top and the outlet is near the bottom because there is a loss of heat from the steam as it passes through the chamber. As the steam loses heat, a part of it condenses back to water and the cold steam gets denser and moves down towards the bottom.

But when done the other way round, the used steam does not pass through the chamber correctly and gets mixed up. This can result in discrepancy in the desired results.

#### Page No 46:

#### Question 6:

When a solid melts or a liquid boils, the temperature does not increase even when heat is supplied. Where does the energy go?

#### Answer:

When a solid melts or a liquid boils, the heat supplied is actually used to break the bond forces between the molecules and bring them apart till the body changes its state completely. Thus, the energy is transferred to the molecules as kinetic energy and the temperature of body remains constant in the process.

#### Page No 46:

#### Question 7:

What is the specific heat capacity of (a) melting ice (b) boiling water?

#### Answer:

(a) The specific heat capacity of melting ice is 0.50 cal/g-^{o}C or 2093 J/kg-K.

(b) The specific heat capacity of boiling water is 0.46 cal/g-^{o}C or 1926 J/kg-K.

#### Page No 46:

#### Question 8:

A person's skin is more severely burnt when put in contact with 1 g of steam at 100°C than when put in contact with 1 g of water at 100°C. Explain

#### Answer:

Steam has greater energy than boiling water due to latent heat of vapourisation. The internal energy of the vapour at 100^{ o}C is greater than the internal energy of the boiling water at the same temperature. Thus, steam burns the skin more severely than boiling water.

#### Page No 46:

#### Question 9:

The atmospheric temperature in the cities on sea-coast change very little. Explain

#### Answer:

The atmospheric temperature in the cities on the sea-coast changes very little due to the following reasons:

1) The specific heat capacity of water is four times greater than land. This means water heats and cools more slowly than land. Also, once the ocean has warmed, it gives up its heat much slower than the land.

2) The moisture content of the air over coastal locations is very high. Moisture in the form of water vapour is the predominate greenhouse gas in the troposphere. Water vapour prevents the loss of heat into space at night. Thus, the temperature during day and night is almost the same.

#### Page No 46:

#### Question 10:

Should a thermometer bulb have large heat capacity or small heat capacity?

#### Answer:

The thermometer bulb should have large heat capacity because if it has lower heat capacity, it may expand at high temperatures resulting in false readings. Large heat capacity of the thermometer bulb ensures the correct reading of temperature by restricting the bulb from expanding.

#### Page No 46:

#### Question 1:

The specific heat capacity of a body depends on

(a) the heat given

(b) the temperature raised

(c) the mass of the body

(d) the material of the body

#### Answer:

(d) the material of the body

Heat capacity of a body is due to their material properties. Due to different molecular structures, different bodies have a different capacity to absorb heat. Therefore, specific heat of a body depends on the material of the body.

#### Page No 46:

#### Question 2:

Water equivalent of a body is measured in

(a) kg

(b) calorie

(c) kelvin

(d) m^{3}

#### Answer:

(a) kg

Since water equivalent of a body is the mass of the water having the same heat capacity as the given body, the water equivalent is measured in kilogram.

#### Page No 46:

#### Question 3:

When a hot liquid is mixed with a cold liquid, the temperature of the mixture

(a) first decreases then becomes constant

(b) first increases then becomes constant

(c) continuously increases

(d) is undefined for some time and then becomes nearly constant

#### Answer:

(d) is undefined for some time and then becomes nearly constant

When hot liquid is mixed with cold liquid, the molecules collide and transfer heat. When heat is transferred, the temperature is undefined. Once the heat energy is shared by the molecules, the system reaches equilibrium and the temperature becomes nearly constant.

#### Page No 46:

#### Question 4:

Which of the following pairs represent units of the same physical quantity?

(a) Kelvin and joule

(b) Kelvin and calorie

(c) Newton and calorie

(d) Joule and calorie

#### Answer:

(d) Joule and calorie

One calorie is defined as the amount of heat needed to raise the temperature of 1 g of water from 14.5^{o} to 15.5^{o} at the pressure of 1 atm. Heat is a form of energy and the unit of energy is joule.

Therefore, joule and calorie represent energy.

#### Page No 46:

#### Question 5:

Which of the following pairs of physical quantities may be represented in the same unit?

(a) Heat and temperature

(b) Temperature and mole

(c) Heat and work

(d) Specific heat and heat

#### Answer:

(c) Heat and Work

As work done in raising temperature of a body is actually the heat supplied to the body, heat and work may be represented in the same unit.

#### Page No 46:

#### Question 6:

Two bodies at different temperatures are mixed in a calorimeter. Which of the following quantities remains conserved?

(a) Sum of the temperatures of the two bodies

(b) Total heat of the two bodies

(c) Total internal energy of the two bodies

(d) Internal energy of each body

#### Answer:

(c) Total internal energy of the two bodies

When two bodies at different temperatures are mixed in the calorimeter, heat flows from one body to the other due to the temperature difference. This results in change in the internal energy of the individual bodies. There is no exchange of heat with the surrounding in the calorimeter. Thus, the total internal energy of the bodies remain conserved as no external work is done on them.

#### Page No 46:

#### Question 7:

The mechanical equivalent of heat

(a) has the same dimension as heat

(b) has the same dimension as work

(c) has the same dimension as energy

(d) is dimensionless

#### Answer:

(d) is dimensionless

If the mechanical work done *(W)* produces the same temperature change as heat *(H),* then the mechanical equivalent of heat (*J) *is equal to *W*/*H. *Thus,

*J* = *W*/*H*

Since the unit of work and heat is the same, mechanical equivalent of heat is dimensionless.

#### Page No 46:

#### Question 1:

The heat capacity of a body depends on

(a) the heat given

(b) the temperature raised

(c) the mass of the body

(d) the material of the body

#### Answer:

(c) the mass of the body

(d) the material of the body

The bigger the body, the larger is its capacity to absorb heat. Therefore, the heat capacity of a body depends on the mass of the body. Also, different bodies have different heat capacities due to their material properties, i.e. due to their molecular structure, the heat capacity of a body depends on the material of the body.

#### Page No 46:

#### Question 2:

The ratio of specific heat capacity to molar heat capacity of a body

(a) is a universal constant

(b) depends on the mass of the body

(c) depends of the molecular weight of the body

(d) is dimensionless

#### Answer:

(c) depends on the molecular weight of the body

Specific heat capacity of a body, $s=\frac{Q}{m\Delta \theta}$

Here,

*Q* = Heat supplied

*m* = Mass of body

Δ*θ** *= Change in temperature

Molar heat capacity of a body, $C=\frac{Q}{n\Delta \theta}\phantom{\rule{0ex}{0ex}}$

Here,

*Q* = Heat supplied

*n* = Number of moles

Δ*θ* = Change in temperature

∴ The ratio of the specific heat capacity and molar heat capacity is given by

$\frac{s}{C}=\frac{\frac{Q}{m\Delta \theta}}{{\displaystyle \frac{Q}{n\Delta \theta}}}=\frac{n}{m}=\frac{n}{nM}=\frac{1}{M}\phantom{\rule{0ex}{0ex}}$

Here,

*M* = Molar mass related to number of moles

*m* = Mass

As the value of *M *is different for different bodies of different composition, the ratio cannot be a universal constant.

Also, the ratio is independent of the mass of the body.

The ratio of the specific heat and molar heat capacity depends on the molecular weight of the body.

Clearly, the unit of molecular weight is kg/mole. So, the ratio that depends only on the molecular weight cannot be dimensionless.

#### Page No 46:

#### Question 3:

If heat is supplied to a solid, its temperature

(a) must increase

(b) may increase

(c) may remain constant

(d) may decrease

#### Answer:

(b) may increase

(c) may remain constant

When heat is supplied to a solid, it is used up either to increase the temperature of the body or to change its state from one form to another by breaking the bonds between the molecules (without raising the temperature).

When heat is supplied to the solid, the internal energy of the solid increases, so the temperature does not decrease.

#### Page No 46:

#### Question 4:

The temperature of a solid object is observed to be constant during a period. In this period

(a) heat may have been supplied to the body

(b) heat may have been extracted from the body

(c) no heat is supplied to the body

(d) no heat is extracted from the body

#### Answer:

(a) heat may have been supplied to the body

(b) heat may have been extracted from the body

If there is no temperature change in a solid object, there is a possibility that the heat might have been supplied to the body that was used up in breaking the bond of the molecules, changing the state of the solid. This is why the temperature of the solid remians constant. Similar is the case when the heat is extracted from the body to change its state.

Since there is a possibility of supplying or extracting heat from the solid, we cannot say that heat is not supplied to the solid or is not extracted from the solid.

#### Page No 46:

#### Question 5:

The temperature of an object is observed to rise in a period. In this period

(a) heat is certainly supplied to it

(b) heat is certainly not supplied to it

(c) heat may have been supplied to it

(d) work may have been done on it

#### Answer:

(c) heat may have been supplied to it

(d) work may have been done on it

If the temperature of an object rises in a period, then there are two possibilities. The heat may have been supplied to it, leading to an increase of the internal energy of the object. That will increase the temperature of the body.

The second possibility is that some work may have been done on it, again leading to an increase of the internal energy of the body. That will also increase the temperature of the body.

#### Page No 46:

#### Question 6:

Heat and work are equivalent. This means,

(a) when we supply heat to a body we do work on it

(b) when we do work on a body we supply heat to it

(c) the temperature of a body can be increased by doing work on it

(d) a body kept at rest may be set into motion along a line by supplying heat to it

#### Answer:

(c) the temperature of a body can be increased by doing work on it

According to the statement "heat and work are equivalent", heat supplied to the body increases its temperature. Similarly, work done on the body also increases its temperature.

For example: If work is done on rubbing the hands against each other, the temperature of the hands increases. So, we can say that heat and work are equivalent.

When heat is supplied to a body, we do not do work on it. When we are doing work on a body, it does not mean we are supplying heat to the body. Also, a body at rest cannot be set in motion along a line by supplying heat to it. So, these statements do not justify the equivalence of heat and work.

#### Page No 47:

#### Question 1:

An aluminium vessel of mass 0.5 kg contains 0.2 kg of water at 20°C. A block of iron of mass 0.2 kg at 100°C is gently put into the water. Find the equilibrium temperature of the mixture. Specific heat capacities of aluminium, iron and water are 910 J kg^{−1} K^{−1}, 470 J kg^{−1} K^{−1} and 4200 J kg^{−1} K^{−1} respectively.

#### Answer:

Given:

Mass of aluminium = 0.5 kg

Mass of water = 0.2 kg

Mass of iron = 0.2 kg

Specific heat of aluminium = 910 J kg^{−1} K^{−1}

Specific heat of iron = 470 J kg^{−1} K^{−1}

Specific heat of water = 4200 J kg^{−1} K^{−1}

Let the equilibrium temperature of the mixture be *T*.

Temperature of aluminium and water = 20°C = 273+20 = 293 K

Temperature of iron = 100°C = 273 + 100 = 373 K

Heat lost by iron, *H*_{1} = 0.2 × 470 × (373 − *T*)

Heat gained by water = 0.2 × 4200 × (*T* − 293)

Heat gained by iron = 0.5 × 910 × (*T*−293)

Total heat gained by water and iron, *H*_{2} = 0.5 × 910 (*T*−293) + 0.2 × 4200 × (*T* − 293)

*H*_{2} = (*T* − 293) [0.5 × 910 + 0.2 × 4200]

We know,

Heat gain = Heat lost

⇒ (*T* − 293) [0.5 × 910 + 0.2 × 4200] = 0.2 × 470 × (373 −*T*)

⇒ (*T* − 293) (455 + 840) = 94 (373 − *T*)

$\Rightarrow \left(T-293\right)\frac{1295}{94}=\left(373-T\right)\phantom{\rule{0ex}{0ex}}\left(T-293\right)\times 14=\left(373-T\right)$

⇒ 14 *T* − 293 × 14 = 373 − *T*

⇒ 15 *T* = 373 + 4102 = 4475

$\Rightarrow T=\frac{4475}{15}=298.33\mathrm{K}\approx 298\mathrm{K}$

∴ *T* = (298 − 273)°C = 25°C

∴ Final temperature = 25° C

#### Page No 47:

#### Question 2:

A piece of iron of mass 100 g is kept inside a furnace for a long time and then put in a calorimeter of water equivalent 10 g containing 240 g of water at 20°C. The mixture attains and equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470 J kg^{−1} °C^{−1}.

#### Answer:

Given:

Mass of iron = 100 g

Water equivalent of calorimeter = 10 g

Mass of water = 240 gm

Let the temperature of surface be $\theta $ °C.

Specific heat capacity of iron = 470 J kg^{−1} °C^{−1}

Total heat gained = Total heat lost

$\Rightarrow \frac{100}{1000}\times 470\times \left(\theta -60\xb0\right)=\frac{\left(240+10\right)}{1000}\times 4200\times \left(60-20\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 47\theta -47\times 60=25\times 42\times 40\phantom{\rule{0ex}{0ex}}\Rightarrow \theta \mathit{}=\frac{42000+2820}{47}=\frac{44820}{47}\phantom{\rule{0ex}{0ex}}=953.61\xb0\mathrm{C}$

#### Page No 47:

#### Question 3:

The temperatures of equal masses of three different liquids A, B and C are 12°C, 19°C and 28°C respectively. The temperature when A and B are mixed is 16°C, and when B and C are mixed, it is 23°C. What will be the temperature when A and C are mixed?

#### Answer:

Given:

Temperature of A = 12°C

Temperature of B = 19°C

Temperature of C = 28°C

Temperature of mixture of A and B = 16°C

Temperature of mixture of B and C = 23°C

Let the mass of the mixtures be *M *and the specific heat capacities of the liquids A, B and C be *C*_{A}, *C*_{B}, and *C*_{C}, respectively.

According to the principle of calorimetry, when A and B are mixed, we get

Heat gained by Liquid A = Heat lost by liquid B

⇒ *MC*_{A} (16 − 12) = *MC*_{B} (19 − 16)

⇒ 4*MC*_{A} = 3 *MC*_{B}

$\Rightarrow M{C}_{\mathrm{A}}=\left(\frac{3}{4}\right)M{C}_{\mathrm{B}}$ ...(1)

When B and C are mixed:

Heat gained by liquid B = Heat lost by liquid C

⇒*MC*_{B} (23 − 19) = *MC*_{C} (28 − 23)

⇒ 4*MC*_{B} = 5 *MC*_{C}

$\Rightarrow {\mathrm{M}}_{\mathrm{CC}}=\left(\frac{4}{5}\right){\mathrm{M}}_{\mathrm{CB}}$ ...(2)

When A and C are mixed:

Let the temperature of the mixture be *T.* Then,

Heat gained by liquid A = Heat lost by liquid C

⇒ *MC*_{A} (*T* − 12) = *MC*_{C} (28 − *T*)

Using the values of *MC*_{A }and *MC*_{C}, we get

$\Rightarrow \left(\frac{3}{4}\right)M{C}_{\mathrm{B}}\left(T-12\right)=\left(\frac{4}{5}\right)M{C}_{\mathrm{B}}\left(28-T\right)\left[\mathrm{From}\mathrm{eqs}.\left(1\right)\mathrm{and}\left(2\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{3}{4}\right)\left(T-12\right)=\left(\frac{4}{5}\right)\left(28-T\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \left(3\times 5\right)\left(T-12\right)=\left(4\times 4\right)\left(28-T\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 15T-180=448-16T\phantom{\rule{0ex}{0ex}}\Rightarrow 31T=628\phantom{\rule{0ex}{0ex}}\Rightarrow T=\frac{628}{31}=20.253\xb0\mathrm{C}\phantom{\rule{0ex}{0ex}}\Rightarrow T=20.3\xb0\mathrm{C}$

#### Page No 47:

#### Question 4:

Four 2 cm × 2 cm × 2 cm cubes of ice are taken out from a refrigerator and are put in 200 ml of a drink at 10°C. (a) Find the temperature of the drink when thermal equilibrium is attained in it. (b) If the ice cubes do not melt completely, find the amount melted. Assume that no heat is lost to the outside of the drink and that the container has negligible heat capacity. Density of ice = 900 kg m^{−3}, density of the drink = 1000 kg m^{−3}, specific heat capacity of the drink = 4200 J kg^{−1} K^{−1}, latent heat of fusion of ice = 3.4 × 10^{5} J kg^{−1}.

#### Answer:

(a)

Given:

Number of ice cubes = 4

Volume of each ice cube = (2 × 2 × 2) = 8 cm^{3}

Density of ice = 900 kg m^{−3}^{ }

Total mass of ice, *m*_{i} = (4×8 ×10^{−6} ×900) = 288×10^{−4} kg

Latent heat of fusion of ice, *L*_{i} = 3.4 × 10^{5} J kg^{−1}

Density of the drink = 1000 kg m^{−3}

Volume of the drink = 200 ml

Mass of the drink = (200×10^{−6})×1000 kg

Let us first check the heat released when temperature of 200 ml changes from 10^{o}C to 0^{o}C.

*H*_{w} = (200×10^{−6})×1000×4200×(10−0) = 8400 J

Heat required to change four 8 cm^{3} ice cubes into water (*H*_{i}) = *m*_{i}*L*_{i} = (288×10^{−4})×(3.4×10^{5}) = 9792 J

Since the heat required for melting the four cubes of the ice is greater than the heat released by water ( *H*_{i} > *H*_{w} ), some ice will remain solid and there will be equilibrium between

ice and water. Thus, the thermal equilibrium will be attained at 0^{o} C.

(b)

Equilibrium temperature of the cube and the drink = 0°C

Let *M* be the mass of melted ice.

Heat released when temperature of 200 ml changes from 10^{o}C to 0^{o}C is given by

*H*_{w} = (200×10^{−6})×1000×4200×(10−0) = 8400 J

Thus,

*M*×(3.4×10^{5}) = 8400 J

Therefore,

*M *= 0.0247 Kg = 25 g

#### Page No 47:

#### Question 5:

Indian style of cooling drinking water is to keep it in a pitcher having porous walls. Water comes to the outer surface very slowly and evaporates. Most of energy needed for evaporation is taken from the water itself and the water is cooled down. Assume that a pitcher contains 10 kg of water and 0.2 g of water comes out per second. Assuming no backward heat transfer from the atmosphere to the water, calculate the time in which the temperature decrease by 5°C. Specific heat capacity of water = 4200 J kg^{−1} °C^{−1} and latent heat of vaporization of water = 2.27 × 10^{6} J kg^{−1}.

#### Answer:

Given:

Specific heat of water ,*S* = 4200 J kg^{−1} °C^{−1}

Latent heat of vapourisation of water ,*L* = 2.27 × 10^{6} J kg^{−1}

Mass,* M =* 0.2 g = 0.0002* *kg

Let us first calculate the amount of energy required to decrease the temperature of 10 kg of water by 5°C.

*U*_{1} = 10 × 4200 J/kg°C × 5°C

*U*_{1} = 210,000 = 21 × 10^{4} J

Let the time in which the temperature is decreased by 5°C be *t.*

Energy required per second for evaporation of water (at the rate of 0.2 g/sec) is given by

*U*_{2}* = ML*

*U*_{2} = (2 × 10^{−4} )× (2.27 × 10^{6}) = 454 J

Total energy required to decrease the temperature of the water = 454 ×* t*

= 21 × 10^{4} J

Now,

*t* $=\frac{21\times {10}^{4}}{454}$ seconds

The time taken in minutes is given by

*t*$=\frac{21\times {10}^{4}}{454\times 60}=7.7\mathrm{minute}s$

∴ The time required to decrease the temperature by 5°C is 7.7 minutes.

#### Page No 47:

#### Question 6:

A cube of iron (density = 8000 kg m^{−3}, specific heat capacity = 470 J kg^{−1} K^{−1}) is heated to a high temperature and is placed on a large block of ice at 0°C. The cube melts the ice below it, displaces the water and sinks. In the final equilibrium position, its upper surface just goes inside the ice. Calculate the initial temperature of the cube. Neglect any loss of heat outside the ice and the cube. The density of ice = 900 kg m^{−3} and the latent heat of fusion of ice = 3.36 × 105 J kg^{−1}.

#### Answer:

Given:

Density of the iron cube = 8000 kg m^{$-$3}

Density of the ice cube = 900 kg m^{$-$3}

Specific heat capacity, *S *= 470 J kg^{−1} K^{−1}

Latent heat of fusion of ice, *L* = 3.36 × 10^{5} J kg^{−1}

Let the volume of the cube be *V*.

Volume of water displaced = *V*

Mass of cube, *m* = 8000 *V *kg

Mass of the ice melted, *M *= 900 *V*

Let the initial temperature of the iron cube be *T* K. Then,

Heat gained by the ice = Heat lost by the iron cube

* m* × *S *× (*T* − 273) = *M × L*

⇒ 8000 *V* × 470 × (*T* − 273) = 900 *V*×( 3.36 × 10^{5})

⇒ 376 × 10^{4} × (*T* − 273) = 3024 × 10^{5}

$\Rightarrow T=\frac{30240+102648}{376}\phantom{\rule{0ex}{0ex}}\Rightarrow T=\frac{132888}{376}\mathrm{K}\phantom{\rule{0ex}{0ex}}\Rightarrow T=353.425\mathrm{K}\approx 353\mathrm{K}\phantom{\rule{0ex}{0ex}}\Rightarrow T=353\mathrm{K}-273\mathrm{K}\phantom{\rule{0ex}{0ex}}\Rightarrow T\mathit{}=80\xb0\mathrm{C}$

#### Page No 47:

#### Question 7:

1 kg of ice at 0°C is mixed with 1 kg of steam at 100°C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = 3.36 × 10^{3} J kg^{−1} and latent heat of vaporization of water = 2.26 × 10^{6} J kg^{−1}.

#### Answer:

Given:

Amount of ice at 0^{o}C = 1 kg

Amount of steam at 100^{o}C = 1 kg

Latent heat of fusion of ice = 3.36 × 10^{3} J kg^{−1}

Latent heat of vapourisation of water = 2.26 × 10^{6} J kg^{−1}

We can observe that the latent heat of fusion of ice (3.36 × 10^{5 }J kg^{−1}) is smaller that latent heat of vapouisation of water (2.26 × 10^{6} ). Therefore, ice will first change into water as less heat is required for this and there will be equilibrium between steam and water.

Heat absorbed by the ice when it changes into water (*Q*_{1}) = 1×(3.36 × 10^{5}) J

Heat absorbed by the water formed to change its temperature from 0^{o}C to 100^{o}C (*Q*_{2}) = 1 × 4200 × 100 = 4.2 × 10^{5} J

Total heat absorbed by the ice to raise the temperature to 100°C, *Q* = *Q*_{1}+*Q*_{2} = 3.36 × 10^{5} + 4.2 × 10^{5} = (3.36 + 4.2) × 10^{5} = 7.56 × 10^{5} J

The heat required to change ice into water at 100^{o}C is supplied by the steam. This heat will be released by the steam and will then change into water.

If all the steam gets converted into water, heat released by steam, *Q'* = 1 ×( 2.26 × 10^{6}) J = 2.26 × 10^{6} J

Amount of heat released is more than that required by the ice to get converted into water at 100^{o}C. Thus,

Extra heat = *Q* − *Q'*

= (2.26 − 0.756) × 10^{6}

= 1.506 × 10^{6}

Let the mass of steam that is condensed into water be *m*. Thus,

*m*$=\frac{7.56\times {10}^{5}}{2.26\times {10}^{6}}$ = 0.335 kg = 335 gm

Total amount of water at 100°C = 1000 + 335 = 1335 g =1.335 g

Steam left = 1− 0.335 = 0.665 kg = 665 gm

#### Page No 47:

#### Question 8:

Calculate the time required to heat 20 kg of water from 10°C to 35°C using an immersion heater rated 1000 W. Assume that 80% of the power input is used to heat the water. Specific heat capacity of water = 42000 J kg^{−1} K^{−1}.

#### Answer:

Given:

Power rating of the immersion rod, *P* = 1000 W

Specific heat of water, *S* = 4200 J kg^{−1} K^{−1}

Mass of water, *M* = 20 kg

Change in temperature, Δ*T* = 25 °C

Total heat required to raise the temperature of 20 kg of water from 10°C to 35°C is given by

*Q *= *M × S *× Δ*T*

*Q *= 20 × 4200 × 25

*Q* = 20 × 4200 × 25 = 21 × 10^{5} J

Let the time taken to heat 20 kg of water from 10°C to 35°C be *t*. Only 80% of the immersion rod's heat is useful for heating water. Thus,

Energy of the immersion rod utilised for heating the water = *t* × (0.80) × 1000 J

*t* × (0.80) × 1000 J = 21 × 10^{5} J

$t=\frac{21\times {10}^{5}}{800}=2625\mathrm{s}\phantom{\rule{0ex}{0ex}}\Rightarrow t=\frac{2625}{60}=43.75\mathrm{min}\approx 44\mathrm{min}$

#### Page No 47:

#### Question 9:

On a winter day the temperature of the tap water is 20°C whereas the room temperature is 5°C. Water is stored in a tank of capacity 0.5 m^{3} for household use. If it were possible to use the heat liberated by the water to lift a 10 kg mass vertically, how high can it be lifted as the water comes to the room temperature? Take g = 10 m s^{−2}.

#### Answer:

Given:

Initial temperature of the water, *T*_{i}_{ }= 20°C

Final temperature of the water (room temperature), *T*_{f}_{ }= 5°C

Change in temeprature, Δ*T* = 20°C − 5°C = 15°C

Volume of water = 0.5 m^{3}

Density of water, d = 1000 kg/m^{3}

Mass of the water,* M* = (0.5 × 1000) kg = 500 kg

Heat liberated as the temperature of water changes from 20°C to 5°C is given by

*Q* = *M×S*×Δ*T*

*Q *= (500 × 4200 × 15) J

*Q *= (500 × 4200 × 15) J

*Q *= (75 × 420 × 1000) J

*Q *= 31500 × 1000 J = 315×10^{5} J

Let the height to which the mass is lifted be *h.*

The energy required to lift the block = *mgh* = 10 × 10 × *h* = 100 *h*

Acording to the question,

*Q = mgh*

$\Rightarrow $100 *h* = 315×10^{5} J

$\Rightarrow $ *h *= 315×10^{3} *m* = 315 km

#### Page No 47:

#### Question 10:

A bullet of mass 20 g enters into a fixed wooden block with a speed of 40 m s^{−1} and stops in it. Find the change in internal energy during the process.

#### Answer:

Given:

Mass of bullet,* m *= 20g = 0.02 kg

Initial velocity of the bullet,* u* = 40 m/s

Final velocity of the bullet = 0 m/s

Initial kinetic energy of the bullet = $\frac{1}{2}m{u}^{2}=\frac{1}{2}\times 0.02\times 40\times 40=16\mathrm{J}$

Final kinetic energy of the bullet = 0

Change in energy of the bullet = 16 J

It is given that the bullet enters the block and stops inside it. The total change in its kinetic energy is responsible for the change in the internal energy of the block.

∴ Change in internal energy of the block = Change in energy of the bullet = 16 J

#### Page No 47:

#### Question 11:

A 50 kg man is running at a speed of 18 km *h*^{−1}. If all the kinetic energy of the man can be used to increase the temperature of water from 20°C to 30°C, how much water can be heated with this energy?

#### Answer:

Given:

Mass of the man, *m *= 50 kg

Speed of the man, *v *= 18 km/h = $18\times \frac{5}{18}=5\mathrm{m}/\mathrm{s}$

Kinetic energy of the man is given by

*K*$=\frac{1}{2}m{\mathrm{V}}^{2}$

$K=\left(\frac{1}{2}\right)50\times {5}^{2}\phantom{\rule{0ex}{0ex}}K=25\times 25=625\mathrm{J}$

Specific heat of the water, *s *= 4200 J/Kg-K

Let the mass of the water heated be *M.*

The amount of heat required to raise the temperature of water from 20°C to 30°C is given by

*Q* = *ms*Δ*T* = *M* × 4200 × (30 − 20)

*Q* = 42000 *M*

According to the question,

*Q = K*

42000 *M* = 625

$\Rightarrow M=\frac{625}{42}\times {10}^{-3}\phantom{\rule{0ex}{0ex}}=14.88\times {10}^{-3}\phantom{\rule{0ex}{0ex}}=15\mathrm{g}$

#### Page No 47:

#### Question 12:

A brick weighing 4.0 kg is dropped into a 1.0 m deep river from a height of 2.0 m. Assuming that 80% of the gravitational potential energy is finally converted into thermal energy, find this thermal energy is calorie.

#### Answer:

Given:

Mass of the brick,* m* = 4 kg

Total vertical distance travelled by the brick, *h* = 3 m

Percentage of gravitational potential energy converted to thermal energy = 80

Total change in potential energy of the brick =* mgh* = 4 × 10 × 3 = 120 J

$\mathrm{Thermal}\mathrm{Energy}=120\times \frac{80}{100}=96\mathrm{J}$

Thermal energy in calories is given by

$U=\frac{96}{4.2}=22.857\mathrm{cal}\approx 23\mathrm{cal}$

#### Page No 47:

#### Question 13:

A van of mass 1500 kg travelling at a speed of 54 km h^{−1} is stopped in 10 s. Assuming that all the mechanical energy lost appears as thermal energy in the brake mechanism, find the average rate of production of thermal energy is cal s^{−1}.

#### Answer:

Given:

Mass of van, *m* = 1500 kg

Speed of van, *v* = 54 km/h

$=54\times \left(\frac{5}{18}\right)=15\mathrm{m}/\mathrm{s}$

Total kinetic energy of the van is given by

$K=\frac{1}{2}m{v}^{2}$

$K=\frac{1}{2}\times 1500\times {\left(15\right)}^{2}\phantom{\rule{0ex}{0ex}}K=750\times 225\phantom{\rule{0ex}{0ex}}K=168750\mathrm{J}\phantom{\rule{0ex}{0ex}}K=\frac{168750}{4.2}\phantom{\rule{0ex}{0ex}}K=40178.57\mathrm{cal}$

Loss in total energy of the van = 40178 cal

Loss in energy per second $=\frac{40178}{10}=4017.8$

$\approx $ 4000 cal./sec

∴ Average rate of production of thermal energy ≈ 4000 cal/sec

#### Page No 47:

#### Question 14:

A block of mass 100 g slides on a rough horizontal surface. If the speed of the block decreases from 10 m s^{−1} to 5 m s^{−1}, find the thermal energy developed in the process.

#### Answer:

Given:

Mass of the block = 100 g = 0.1 kg

Initial speed of the block = 10 m/s

Final speed of the block = 5 m/s

Initial kinetic energy of the block = $\frac{1}{2}\times 0.1\times {10}^{2}=5\mathrm{J}$

Final kinetic energy of the block = $\frac{1}{2}\times 0.1\times {5}^{2}=1.25\mathrm{J}$

Change in kinetic energy of the block = 5 $-$ 1.25 = 3.75 J

Thermal energy developed is equal to the change in kinetic energy of the block. Thus,

Thermal energy developed in the process = 3.75 J

#### Page No 47:

#### Question 15:

The blocks of masses 10 kg and 20 kg moving at speeds of 10 m s^{−1} and 20 m s^{−1} respectively in opposite directions, approach each other and collide. If the collision is completely inelastic, find the thermal energy developed in the process.

#### Answer:

Given:

Mass of the first block, *m*_{1} = 10 kg

Mass of the second block, *m*_{2} = 20 kg

Initial velocity of the first block, *u*_{1} = 10 m/s

Initial velocity of the second block, *u*_{2} = 20 m/s

Let the velocity of the blocks after collision be *v*.

Applying conservation of momentum, we get

* **m*_{2}*u*_{2} − *m*_{1}*u*_{1} = (*m*_{1} + *m*_{2})*v*

⇒ 20 × 20 − 10 × 10 = (10 + 20)*v*

⇒ 400 − 100 = 30 *v*

⇒ 300 = 30 *v*

⇒ *v* = 10 m/s

Initial kinetic energy is given by

${K}_{\mathrm{i}}=\frac{1}{2}{m}_{1}{u}_{1}^{2}+\frac{1}{2}{m}_{2}{u}_{2}^{2}$

${K}_{\mathrm{i}}=\frac{1}{2}\times 10\times {\left(10\right)}^{2}+\frac{1}{2}\times 20\times {\left(20\right)}^{2}\phantom{\rule{0ex}{0ex}}{K}_{\mathrm{i}}=500+4000=4500$

Final kinetic energy is given by

${K}_{\mathrm{f}}\mathit{=}\frac{1}{2}\left({m}_{1}\mathit{+}{m}_{2}\right){v}^{2}$

${K}_{\mathrm{f}}=\frac{1}{2}\left(10+20\right){\left(10\right)}^{2}\phantom{\rule{0ex}{0ex}}{K}_{\mathrm{f}}=\left(\frac{30}{2}\right)\times 100=1500$

∴ Total change in KE = 4500 J − 1500 J = 3000 J

Thermal energy developed in the process = 3000 J

#### Page No 47:

#### Question 16:

A ball is dropped on a floor from a height of 2.0 m. After the collision it rises up to a height of 1.5 m. Assume that 40% of the mechanical energy lost goes as thermal energy into the ball. Calculate the rise in the temperature of the ball in the collision. Heat capacity of the ball is 800 J K^{−1}.

#### Answer:

Height of the floor from which ball is dropped, *h*_{1} = 2.0 m

Height to which the ball rises after collision, *h*_{2} = 1.5 m

Let the mass of ball be *m* kg.

Let the speed of the ball when it falls from *h*_{1} and *h*_{2} be *v*_{1} *and **v*_{2}, respectively.

${v}_{1}=\sqrt{2g{h}_{1}}=\sqrt{2\times 10\times 2}=\sqrt{40}\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}{v}_{2}=\sqrt{2g{h}_{2}}=\sqrt{2\times 10\times 1.5}=\sqrt{30}\mathrm{m}/\mathrm{s}$

Change in kinetic energy is given by

$\u2206K=\frac{1}{2}\times \mathrm{m}\times 40-\left(\frac{1}{2}\mathrm{m}\right)\times 30=\left(\frac{10}{2}\right)\mathrm{m}\phantom{\rule{0ex}{0ex}}\Rightarrow \u2206K=5\mathrm{m}$

If the position of the ball is considered just before hitting the ground and after its first collision, then 40% of the change in its KE will give the change in thermal energy of the ball. At these positions, the PE of the ball is same. Thus,

Loss in PE = 0

The change in kinetic energy is utilised in increasing the temperature of the ball.

Let the change in temperature be Δ*T**.* Then,

$\left(\frac{40}{100}\right)\times \u2206K=m\times 800\times \u2206T\phantom{\rule{0ex}{0ex}}\left(\frac{40}{100}\right)\times \frac{10}{2}m=m\times 800\times \u2206T\phantom{\rule{0ex}{0ex}}\Rightarrow \u2206T=\frac{1}{400}=0.0025\phantom{\rule{0ex}{0ex}}=2.5\times {10}^{-3}\xb0\mathrm{C}$

#### Page No 47:

#### Question 17:

A copper cube of mass 200 g slides down on a rough inclined plane of inclination 37° at a constant speed. Assume that any loss in mechanical energy goes into the copper block as thermal energy. Find the increase in the temperature of the block as it slides down through 60 cm. Specific heat capacity of copper = 420 J kg^{−1} K^{−1}.

#### Answer:

Mass of copper cube, *m* = 200 g = 0.2 kg

Length through which the block has slided, *l* = 60 cm = 0.6 m

Since the block is moving with constant velocity, the net force on it is zero. Thus,

Force of friction, *f* = mg

Also, since the object is moving with a constant velocity, change in its K.E will be zero.As the object slides down, its PE decreases at the cost of increase in thermal energy of copper.

The loss in mechanical energy of the copper block = Work done by the frictional force on the copper block to a distanceof 60 cm

*W* = *mg* *l* sin *θ*

*W* = 0.2 × 10 × 0.6 sin 37°

*W* $=1.2\times \left(\frac{3}{5}\right)=0.72$

Let the change in temperature of the block be ∆*T*.

Thermal energy gained by block = *ms* ∆*T* = 0.2 × 420×∆*T* = 84∆*T*

But 84∆*T* = 0.72

$\Rightarrow \u2206T=\frac{0.72}{84}=0.00857\phantom{\rule{0ex}{0ex}}\u2206T=0.0086=8.6\times {10}^{-3}\xb0\mathrm{C}$

#### Page No 47:

#### Question 18:

A metal block of density 600 kg m^{−3} and mass 1.2 kg is suspended through a spring of spring constant 200 N m^{−1}. The spring-block system is dipped in water kept in a vessel. The water has a mass of 260 g and the bloc is at a height 40 cm above the bottom of the vessel. If the support of the spring is broken, what will be the rise in the temperature of the water. Specific heat capacity of the block is 250 J kg^{−3} K^{−1} and that of water is 4200 J kg^{−1} K^{−1}. Heat capacities of the vessel and the spring are negligible.

#### Answer:

Given:

Density of metal block, *d* = 600 kg m^{−3}

Mass of metal block, *m* = 1.2 kg

Spring constant of the spring,* k* = 200 N m^{−1}

Volume of the block, *V*$=\frac{1.2}{6000}=2\times {10}^{-4}{\mathrm{m}}^{3}$

When the mass is dipped in water, it experiences a buoyant force and in the spring there is potential energy stored in it.

If the net force on the block is zero before breaking of the support of the spring, then

*kx* + *V*ρ*g* = *mg*

200*x* + (2 × 10^{−4})× (1000) × (10) = 12

$\Rightarrow x=\frac{\left(12-2\right)}{200}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{10}{200}=0.05\mathrm{m}$

The mechanical energy of the block is transferred to both block and water. Let the rise in temperature of the block and the water be Δ*T*.

Applying conservation of energy, we get

$\frac{1}{2}k{x}^{2}+mgh-V\mathrm{\rho}gh={m}_{1}{s}_{1}\u2206T+{m}_{2}{s}_{2}\u2206T\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\times 200\times 0.0025+1.2\times 10\times \left(\frac{40}{100}\right)-2\times {10}^{-4}\times 1000\times 10\times \left(\frac{40}{100}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{260}{1000}\right)\times 4200\times \u2206T+1.2\times 250\times \u2206T\phantom{\rule{0ex}{0ex}}\Rightarrow 0.25+4.8-0.8=1092\u2206T+300\u2206T\phantom{\rule{0ex}{0ex}}\Rightarrow 1392\u2206T=4.25\phantom{\rule{0ex}{0ex}}\Rightarrow \u2206T=\frac{4.25}{1392}=0.0030531\phantom{\rule{0ex}{0ex}}\u2206T=3\times {10}^{-3}\xb0\mathrm{C}$

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