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 in the vessel. It is given that the specific heats of water and ice are 1 $kcal$ ${kg}^{-1}$ ${}^{\circ }{C}^{-1}$ and 0.5 kcal ${kg}^{-1}$ ${}^{\circ }{C}^{-1}$ respectively and the latent heat of fusion of ice is 80 $kcal/kg$.

Answer 1

Given: 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. It is given that the specific heats of water and ice are $1kcal/\left(k{g}^{\circ }C\right)$ and $0.5kcal/\left(k{g}^{\circ }C\right)$ respectively and the latent heat of fusion of ice is $80kcal/kg$.

To find the final mass of water in the vessel.

Solution:

As per the given criteria,

Mass of water, $m_w=5kg$

Mass of the ice, $m_i=2kg$

Temperature of ice, $T_i=-{20}^{\circ }C$

Temperature of water, $T_w={20}^{\circ }C$

Specific heat of ice, $s_i=0.5kcal/\left(k{g}^{\circ }C\right)$

Specific heat of water, $s_w=1kcal/\left(k{g}^{\circ }C\right)$

latent heat of fusion of ice is $L=80kcal/kg$.

Let $M$ be the final mass of the content

The amount of heat losed by water at 0 degreeis

$Q_w=m_w{s}_s\left(\Delta T\right)=5\times 1\times 20=100kcal$

the amount of heat gained by ice to go to 0 degree is

$Q_i=m_i{s}_i\left(\Delta T\right)=2\times 0.5\times 20=20kcal$

heat left for absorption is

$Q_w-Q_i=100-20=80kcal$

This is equal to latent heat, i.e.,

$mL=80⟹m=\frac{80}{L}=\frac{80}{80}=1kg$

This is the mass of water from the converted ice.

So the final mass of water in the vessel is , $M=m_w+m=5+1=6kg$