Question

2 kg of ice at $-{20}^{o}C$ is mixed with 5 kg of water at ${20}^{o}C$ in an insulating vessel having a negligible heat capacity. Calculate the final mass of water in the vessel. It is given that the specific heats of water and ice are $1kcalk{g}^{-1}{}^o{C}^{-1}$ and $0.5kcalk{g}^{-1}{}^o{C}^{-1}$ respectively and the latent heat of fusion of ice is 80 kcal/kg.

• A. 5 kg
• B. 6 kg
• C. 7 kg
• D. 8 kg

Answer 1

The correct option is B 6 kg

Heat lost by a 5 kg of water to reduce its temperature to $0^{o}C\left(H_L\right)$
$0.5\times 1000\times (20-0)$
$=10\times {10}^{4}cal$
Heat gained by 2 kg ice to increase its temperature from $-{20}^{o}C$ to $0^{o}C$
$H_1=2\times 0.5\times {10}^3\times \left(20\right)$
$=20\times {10}^{3}cal=2\times {10}^{4}cal$
Let mass of ice melted $m_{ice}$
Then amount of heat supplied to 2 kg ice at $0^{o}C=H_L-H_l$
$=(10\times {20}^4-2\times {10}^4)cal$
$=8\times {10}^{4}cal$
$\therefore 8\times {10}^4=m_{ice}\times 80\times {10}^3$
$m_{ice}=1kg$
$\therefore$ Final mass of the water in the vessel $=(5+1)=6kg$.