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Heat

Heat: a form of energy and is measurable

Take the same amount of water in two similar beakers. Put a thermometer in each beaker, and heat the beakers. Heat one beaker for 5 minutes and the other for 10 minutes. What do you observe?

You will observe that the water in the beaker that is heated for the longer time becomes hotter than the other beaker. The thermometer in this beaker shows a higher temperature. Why does this happen?

The answer is very simple—the longer the time we heat water, the hotter it becomes. So, we can say the rise in the temperature of a body is directly proportional to the heat energy contained in the body.

Let us do an experiment to explore more properties of heat.

As shown in the figure, take two beakers: one with 100 mL of water and the other with 200 mL of water. Now, start heating the beakers, and heat them till the temperatures in both the beakers become the same. Do they require the same time?

You will observe that the time taken to raise the temperature of water by 30°C in the 200-mL beaker is double the time taken to raise the temperature of water (by the same measure) in the 100-mL beaker. Perform this experiment a few more times by raising the temperature to different levels.

Every time you will get the same result.

Hence, we can conclude that the amount of heat absorbed by a body to raise its temperature is directly proportional to the mass of the body. The more the mass of a body, the more is the heat required by it to raise its temperature.

Again, we take the same two beakers. Fill the beaker containing 200 mL of water with 100 mL of oil instead.

Heat the two beakers for the same amount of time. You will observe that the temperature of oil increases more than that of water for the same amount of heat energy supplied. Thus, it can be concluded that water needs more heat energy than oil to be raised to the same degree of temperature.

Hence, it can be concluded that heat energy contained in a body is directly related to the chemical composition of the body.

Therefore, from all the experiments we can conclude that

• When the mass (m) is constant, the rise in the temperature (Δθr ) of a body is directly proportional to the heat energy (H) supplied to it, i.e.,

H ∝Δθr(mis constant)

• When the change in temperature is constant, the heat energy (H) absorbed by an object is directly proportional to the mass (m) of the object, i.e.,

H m(Δθris constant)

• When both mass (m) and temperature change (Δθr ) are constant, the heat energy (H) absorbed by an object depends on its chemical composition.
• Therefore, we can say that

H mΔθr(Both m andΔθr are variable)

H = mCΔθr

Where, C is a proportionality constant, called specific heat capacity. This depends on the nature of the substance.

Thus,

Heat absorbed = Mass of the body × Specific heat capacity × Rise in temperature

Units of Specific heat capacity

Specific capacity (C) =

In the CGS system, the unit of heat is dyne centimeter (dyn-cm); that of mass is gram (g), and that of temperature is °C.

Therefore, the unit of specific heat capacity in the CGS system is (dyn-cm g−1 °C−1)

In the SI system, the unit of heat is joule (J); that of mass is kilogram (kg), and that of temperature is Kelvin (K).

Therefore, the unit of specific heat capacity in the SI system J kg−1 K−1

Water has the highest specific heat capacity of 4.2 J/g °C. In other words, every 1 g of water will absorb 4.2 J of energy when its temperature rises by 1°C.

Measurement of Specific Heat

Measurement of specific heat of an object includes two components: mixing method and calorimeter. In this method,
a hot solid object is put in the water in a calorimeter. As soon as the hot object is placed in water, exchange of heat occurs between the hot object and water in the calorimeter. This exchange of heat continues till the temperatures of the solid object, water and the calorimeter become equal.
So, we can say that

Heat lost by solid object = Heat gained by water in calorimeter + Heat gained by the calorimeter     .....(1)

The following equation will give us an idea about how to calulate the varoius components which are present in the above equation.

Heat lost by the solid object (Q) = Mass of the solid object $×$ Its specific heat $×$ Decrease in its temperature     .....(2)
Heat gained by the water (Q1) = Mass of the water $×$ Its specific heat $×$ Increase in its temperature     .....(3)
Heat gained by the calorimeter (Q2) = Mass of the calorimeter $×$ Its specific heat $×$ Increase in its temperature     .....(4)

Using equation 2, 3 and 4, we can rewrite equa…

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