19 g of water at 30⁰C and 5 g of ice at $-$20⁰C are mixed together in a calorimeter. What is the final temperature of the mixture? (Take, specific heat of ice = 0.5 cal ${\mathrm{g}}^{-1}$ ⁰$\mathrm{C}{ }^{-1}$ and latent heat of fusion of ice = 80 cal ${\mathrm{g}}^{-1}$)

Here, specific heat of ice,  $c_{\text{ice }}=0.5{\mathrm{calg}}^{-1\mathrm{o}}{\mathrm{C}}^{-1}$

Specific heat of water, $c_{\text{water }}=1{\mathrm{calg}}^{-1\mathrm{o}}{\mathrm{C}}^{-1}$

Latent heat of fusion of ice ,$L_{\text{ice }}=80{\mathrm{calg}}^{-1}$

Here, ice will absorb heat while hot water will release it.
Let T be the final temperature of the mixture.
Assuming water equivalent of calorimeter to be neglected.

Heat given by water,   $Q_1=m_{\text{water }}{c}_{\text{water }}\mathrm{\Delta }T$

$=19\times 1\times (30-T)=570-19T\dots \left(\mathrm{i}\right)$

Heat absorbed by ice,

$Q_2=m_{\text{ice }}\times c_{\text{ice }}\times [0-(-20\left)\right]+m_{\text{ice }}\times L_{f\text{ ice }}$ $+m_{\text{ice }}\times c_{\text{water }}\times (T-0)$

$=5\times 0.5\times 20+5\times 80+5\times 1\times T=450+5T\dots \text{ (ii) }$

According to principle of calorimetry,  $Q_1=Q_2$

$570-19T=450+5T\Rightarrow T=\frac{120}{24}=5^{\circ }\mathrm{C}$