Question

Derive an expression for finding out the specific heat capacity of a body (solid) from the readings of an experiment given below:

(i) Mass of empty calorimeter (with stirrer) = m₁ gm

(ii) Mass of the metal piece = M gm

(iii) Mass of colorimeter and water = m2 gm

(iv) Initial temperature and water = t₁°C

(v) Temperature of hot solid (metal piece) = t₂ °C

(vi) Final temperature of the mixture = t°C

(vii) Specific heat of calorimeter = 0.4 J gm / °C

Answer 1

Lets J gm / °C be the specific heat of the solid

(i) Fall in temperature of hot solid = (t₂ - t) °C 

(ii) Rise in the temperature of calorimeter as well as water = (t - t₂) °C

(iii) Heat taken up by calorimeter = m₁ × 0.4 × (t - t₁) °C 

(iv) Heat taken up by water = (m₂ - m₁) × 4.2 × (t - t₁)°C 

(v) Heat given out by hot soild = M × s × (t₂ - t)°C 

Now, heat given out by hot solid = Heat taken up by calorimeter + Heat taken up by water

or Ms (t₂ - t) = m₁ × 0.4 (t - t₁) + (m₂ - m₁) × (4.2 × (t - t₁))

From this heat equation, 's' can b e calculated where all other values are known.