Question

A piece of ice of mass $40g$ is added to $200g$ of water at ${50}^{0}C$. Calculate the final temperature of water when all the ice has melted. Specific heat capacity of water $=4200Jk{g}^{-1}{K}^{-1}$ and specific latent heat of fusion of ice $=336\times {10}^{3}Jk{g}^{-1}$.

Answer 1

Let final temperature of water when all the ice has melted = T°C.

Amount of heat lost when 200g of water at 50°C cools to T°C =200 ×4.2 ×(50 − T) = 42000 − 840T

Amount of heat gained when 40g of ice at 0oC converts into water at 0°C.= 40 ×336J = 13440 J

Amount of heat gained when temperature of 40g of water at 0°C rises to T°C = 40 ×4.2 ×(T-0) = 168T

We know thatAmount of heat gained = amount of heat energy lost.

13440 + 168T = 42000 − 840T

= 42000 − 13440 = 1008T

$T$ = 28560/1008 = 28.33° C.