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 the 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. $5.5\text{kg}$
• B. $6\text{kg}$
• C. $5\text{kg}$
• D. $7\text{kg}$

Answer 1

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

The heat energy absorbed, when $2\text{kg}$ of ice at $-{20}^{\circ }\text{C}$ is heated to ice at $0^{\circ }\text{C}$:
$Q_{ice}=m{S}_{ice}\Delta T$
$=2\times 0.5\times 20=20\text{kcal}$

The heat energy released, when $5\text{kg}$ of water at ${20}^{\circ }\text{C}$ is cooled to water at $0^{\circ }\text{C}$:
$Q_{water}=m{S}_{water}\Delta T$
$=5\times 1\times 20=100\text{kcal}$

Thus, the energy available to melt the ice is:
$Q=Q_{water}-Q_{ice}=100-20=80\text{kcal}$

The mass of ice melted by this energy is:
$m=\frac{Q}{L_{fusion}}=\frac{80}{80}=1\text{kg}$ (remaining mass of the ice will not melt)
Hence, the final mass of water in the container is:
$m_w+m=5+1=6\text{kg}$