# Specific Heat Capacity of Liquid by Method of Cooling | Physics Grade XI

### Heat Capacity

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Introduction to Kinematics

# Specific Heat Capacity of Liquid by Method of Cooling

Specific Heat Capacity of a Liquid by the Method of Cooling
When two liquids are cooled under identical conditions their rates of cooling are equal.

Take two identical calorimeters A and B of the same material having masses m1 and m­2 respectively as shown in the figure. let s be the specific heat capacity of the calorimeters. Let the calorimeter A be filled with water of mass M1 and the calorimeter B be filled with the experimental liquid of the same volume of mass M2. The initial temperature of the two liquids is noted as θ1. Let t1 and t2 be the amount of time taken by the liquid to cool from θ1 to θ2.

Heat lost by the calorimeter A = M1s (θ1 – θ2)
And heat lost by the water in calorimeter A = m1s11 – θ2)
Where s1 is the specific heat capacity of water.

Therefore, total heat lost by the calorimeter A and water in cooling from θ1 to θ2;
= M1s (θ1 – θ2) + m1s11 – θ2)
= (θ1 – θ2) (M1s + m1s1) …..... (i)
The rate of cooling of water and calorimeter = (θ1 – θ2) (M1s + m1s1)/ t1 …...... (ii)

Similarly, the rate of cooling of liquid and calorimeter B is;
= (M2s + m2s') (θ1 – θ2)/ t2 …...... (iii)
Where s' is the specific heat capacity of liquid

From the law of cooling, we have
Rate of cooling of water and calorimeter A = rate of cooling of liquid and calorimeter B
or, (θ1 – θ2) (M1s + m1s1)/ t1= (M2s + m2s') (θ1 – θ2)/ t2
or, s' = [(m1s + M1s1)/ m2] * (t2/ t1­) – (M2/ m2) * s ....… (iv)

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