Question

A piece of ice of mass 40 g is added to 200 g of water at ${50}^{o}C$. Calculate the final temperature of water when all the ice has melted.Specific heat capacity of water=4200 J k$g^{-1}{K}^{-1}$ and specific latent heat of fusion of ice =336 $\times {10}^{3}Jk{g}^{-1}$.

Answer 1

Let the final temperature of water be T°C.

Now, heat gained by ice = heat lost by water

$\Rightarrow 40\times 336+40\times 4.2\times (T-0)=200\times 4.2\times (50-T)$

$\Rightarrow 13440+168T=42000-840T$

$\Rightarrow 1008T=28560$

$\therefore T={28.33}^{o}C$