Question

Ice at $0^{\circ }C$ is added to 200g of water initially at ${70}^{\circ }C$ in a vacuum flask. When 50g of ice has been added and has all melted the temperature of the flask and contents is ${40}^{\circ }C$. When a further 80g of ice has been added and has all melted the temperature of the whole becomes ${10}^{\circ }C$. Find the latent heat of fusion of ice in cal/g.

Answer 1

The specific heat capacity of water is $1cal/gm{/}^{\circ }C$ and for the ice is $0.5cal/gm{/}^{\circ }C$.

When the $50gms$ of ice is added to $200gms$ of water, the latent heat is given as,

$L_1=\frac{C_w{m}_w\Delta T_w-C_w{m}_i\Delta T_i}{m_i}$

$=\frac{1\times 200\times \left(30\right)-1\times 50\times 40}{50}$

$=80cal/gm$

When the $80gms$ of ice is added to $250gms$ of water, the latent heat is given as,

$L_2=\frac{1\times 250\times \left(30\right)-1\times 80\times 10}{80}$

$=83.75cal/gm$

Thus, when the temperature of the water is lowered to $10{}^{\circ }C$, the latent heat of fusion of ice will be $80cal/gm$.