Question

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

• A. $7\text{kg}$
• B. $6\text{kg}$
• C. $4\text{kg}$
• D. $2\text{kg}$

Answer 1

The correct option is B $6\text{kg}$

Heat released by $5\text{kg}$ of water when its temperature falls from ${20}^{\circ }\text{C}$ to $0^{\circ }\text{C}$ is,
$Q_1=mc\Delta \theta =\left(5\right)\left({10}^3\right)(20-0)={10}^5\text{cal}$
when $2\text{kg}$ ice at $-{20}^{\circ }\text{C}$ comes to a temperature of $0^{\circ }\text{C}$, it takes an energy
$Q_2=mc\Delta \theta =\left(2\right)\left(500\right)\left(20\right)=0.2\times {10}^5\text{cal}$
the remaining heat
$Q=Q_1-Q_2=0.8\times {10}^5\text{cal}$ will melt a mass $m$ of the ice, where
$m=\frac{Q}{L}=\frac{0.8\times {10}^5}{80\times {10}^3}=1\text{kg}$
So, the temperature of the mixture will be $0^{\circ }C$, mass of water in it is $5+1=6\text{kg}$ and mass of ice is $2-1=1\text{kg}$