Question

A piece of iron weighing $15g$ is dropped into a cavity in a block of ice at $0^{\circ }$ . As a result, $2.5g$ of ice get melted. It the temperature of iron was ${113.6}^{\circ }C$, calculate the sp. Heat of iron. ($L$ for fusion of ice $=80cal/g)$.

Answer 1

Remember one principle of calorimetry.

Heat lost by hot body = heat gained by cold body unless any heat is lost to the surroundings.

Whenever any solid object melts, some heat is absorbed only for melting into liquid. The temperature of the object doesn't rise in such a situation unless it further absorbs any heat after melting. This heat absorbed is known as latent heat of melting.

In the problem, since the incidence is a quick one, we can assume that negligible heat is lost to the surroundings and the temperature of the ice-cold water i.e. the melted ice doesn't rise.

Heat lost by iron = change in temperature$\times$mass of iron$\times \eta$

Here $\eta$ is the specific heat of iron.

Heat gained by ice=mass of ice melted$\times$latent heat of melting of ice

$\therefore (113.6-0)\times 15\times \eta =2.5\times 80$

$\therefore \eta =\frac{2.5\times 80}{113.6\times 15}=0.12$

Answer:- $0.12cal/g^{o}C$