Question

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

• A. 7 kg
• B. 6 kg
• C. 4 kg
• D. 2 kg

Answer 1

The correct option is B 6 kg

Heat lost by water at ${20}^{\circ }C=m{s}_W\Delta T$
$H=5\times 1\times 20=100kcal$
At first, ice at $(-{20}^{\circ }C)$ will take heat to change into ice at $0^{\circ }C$.
$H=m{s}_{ice}\Delta T=\left(2kg\right)\times 0.5\times 20=20kcal$
$\therefore$ After this, heat available = $(100-20)=850kcal$
This heat will now be gained by ice at $0^{\circ }C$ to melt into water at $0^{\circ }C$. Let m kg of ice melt.
$\therefore m\times 80=80\therefore m=1kg$
Out of 2 kg of ice, 1 kg of ice melts into water and 1 kg of ice remains unmelted in container.
$\therefore$ Amount of water in container = 5 + 1 = 6kg