Question

A piece of ice of mass $40g$ is added to $200g$ of water at ${50}^{o}C$. 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 $=336\times {10}^{3}Jk{g}^{-1}$

• A. ${28.67}^{o}C$
• B. ${21.67}^{o}C$
• C. ${28.33}^{o}C$
• D. ${21.33}^{o}C$

Answer 1

The correct option is C ${28.33}^{o}C$

For water, $m_1=200,c_1=4200,t_1={50}^{o}C$

For Ice, $m_2=40,c_2=336000$

Heat required to melt ice is $mass\times Latentheatoffusion=40\times {10}^{-3}\times 336\times {10}^3=13440J$.

Let final temperature be $T$,

For water, $m=200\times {10}^{-3},c=4200,t=50$

So energy supplied by water is $4200\times 200\times {10}^{-3}\times (50-T)$

After ice is melted to water, $m=40\times {10}^{-3},c=4200,t=0$

So energy absorbed is $4200\times 40\times {10}^{-3}\times T$

Balancing energy,

$4200\times 200\times {10}^{-3}\times (50-T)$=13440+$4200\times 40\times {10}^{-3}\times T$

Solving above, $T={28.33}^{o}C$