Question

A piece of ice of mass $40g$ at $0^{o}C$ is added to $200g$ of water at ${50}^{\circ }C$. 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. $28.33K$
• B. ${28.33}^{\circ }C$
• C. ${2.833}^{\circ }C$
• D. ${28.67}^{\circ }C$

Answer 1

The correct option is B ${28.33}^{\circ }C$

Let the final temperature be $T$

ice$\left(0^{o}C\right)$ is $40g(40\times {10}^{-3}Kg)$

water$\left({50}^{o}C\right)$ is $200g(200\times {10}^{-3}Kg)$

$CASE-I$

Heat gained by ice at $0^{o}C$ from converting to water at $0^{o}C=mL=40\times {10}^{-3}\times 336\times {10}^{3}J$

Also heat gain by water at $0^{o}C$ to achieve final temperature $T=ms△T=40\times {10}^{-3}\times 4200\times (T-0)$

$\therefore$ Total heat gained by ice at $0^{o}C$ to change water at $T^{o}C=mL+ms△T=(4\times 42T)+(40\times 336)$

$=13440+168T\left(1\right)$

$CASE-II$

Heat lost by water as its temperature decreases from ${50}^{o}C$ to $T^{o}C=0.2\times \left(4200\right)(50-T)$

$=4200-840T\left(2\right)$

Heat gain by ice = Heat lost by water $\left(\right(1)=(2\left)\right)$

$\Rightarrow 13440+168T=4200-840T\Rightarrow T={28.3}^{o}C$