Heat , Specific Heat Capacity , Molar Heat Capacity , Latent Heat , Water-Equivalent , Principle of Calorimetry

Heat : When two bodies at different temperature are placed in contact , the energy is transferred from higher temperature to lower temperature known as Heat .

Heat & Work equivalent :

Whenever mechanical work is transformed into heat or heat into mechanical work, there is a constant ratio between the work and the amount of heat. This ratio is called “mechanical equivalent of heat” and is denoted by J .

Thus, if W be the amount of work done and Q the amount of heat produced, we have

W/Q = J

W = JQ

If Q = 1 unit then J = W . Therefore, J is numerically equal to the mechanical work required to produce one unit of heat.

Where , J = Mechanical equivalent of heat

J = 4.2 Joule/cal


The amount of heat needed to increase the temperature of 1 gm of water from 14.5°C to 15.5°C at STP is known as 1 calorie

Specific Heat Capacity:

It is heat required to raise temperature by 1° C or 1° K for unit mass of the body.

$\large s = \frac{\Delta Q}{m \Delta T} $

$\large \Delta Q = m s \Delta T $

$ \displaystyle dQ = m s dT $

$ \displaystyle Q = m\int_{T_1}^{T_2} s dT $
(be careful about unit of temperature, use units according to the given units of s )

Latent Heat:

The amount of heat required to change one phase of 1 gm of a substance to another phase.

Q = m L

L = latent heat of substance in cal/gm or in Kcal/kg.

Lice = 80 cal/gm – Specific latent heat of fusion of ice

Lsteam = 540 cal/gm – Specific latent heat of vaporization of water

Molar Heat Capacity:

If instead of unit mass we consider one mole of a substance, the heat required to change the temperature of one mole of a substance through 1°C (or K) is called molar heat capacity or molar specific heat and is represented by C.  If the molecular weight of a substance is M :

$ \displaystyle C = \frac{Q}{n \Delta T} $

As , $ \displaystyle s = \frac{Q}{m \Delta T} \quad and \quad n = \frac{m}{M}$

Its SI units are (J/mol K)

Thermal-capacity :

If instead of unit mass we consider the whole body, (of mass m), the heat required to raise the temperature of a given body by 1°C is called its thermal capacity , i.e.,

Thermal capacity = $ \displaystyle m s = n C = \frac{Q}{\Delta T} $

Thermal capacity of a body depends on the mass and nature of body. It has units (J/K) or cal/°C

Water-Equivalent :

If thermal capacity of a body is expressed in terms of mass of water it is called water-equivalent of the body,

i.e., water-equivalent of a body is the mass of water which when given same amount of heat as to the body, changes the temperature of water through same range as that of the body, i.e.,

$ \displaystyle W = (m \times s ) $ gram

[where : W = mass of water ; m = mass of substance ]

The unit of water equivalent W is g while its dimension [M].

Principle of Calorimetry

When two bodies (one being solid and other liquid or both being liquid) at different temperature are mixed, heat will be transferred from body at higher temperature to a body at lower temperature till both acquire same temperature.

The body at higher temperature releases heat while body at lower temperature absorbs it, so that :

Heat lost = Heat gained

i.e. principle of calorimetry represents the law of conservation of heat energy.

Solved Example : The temperature of equal masses of three different liquids A, B and 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 is 23° C. What would be the temperature when A and C are mixed ?

Solution: In accordance with principle of calorimetry :

When A and B are mixed

m SA (16 – 12) = m SB (19 – 16)

⇒ SA = (3/4)SB

and when B and C are mixed

m SB (23 – 19) = m SC (28 – 23)

SC = (4/5)SB

Now when A and C are mixed if T is the common temperature of mixture:

m SA (T – 12) = m SC (28 – T)

Substituting SA and Sc from above,

(3/4)(T – 12) = (4/5)(28 – T)

which on solving gives,

T = 20.25 ° C

Solved Example : A solid material is supplied heat at a constant rate. The temperature of the material is changing with the heat input as shown in figure. Study the graph carefully and answer the following questions :

(i) What do the horizontal regions AB and CD represent?

(ii) If CD = 2 BA , what do you infer?

(iii) What does slope DE represent?

(iv) The slope of OA > the slope of BC .What does this indicate ?

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