Question

$2\text{kg}$ of ice at $-{20}^{\circ }C$ is mixed with $5\text{kg}$ 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 $1\text{kcal/kg}$ per ${}^{\circ }C$ and $0.5\text{kcal/kg}$ per ${}^{\circ }C$ 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 $6\text{kg}$

Initially ice will absorb heat to raise its temperature to $0^{\circ }C$, then its melting takes places.
If $m_1$ = Initial mass of ice, $m_1^{\prime }$ = Mass of ice that melts and $m_w$ = Initial mass of water
By law of mixture,
heat gained by ice = heat lost by water
$\Rightarrow m_1\times c\times \left(20\right)+m_1^{\prime }\times L=m_w{c}_w\left(20\right)$
$\Rightarrow 2\times 0.5\left(20\right)+m_1^{\prime }\times 80=5\times 1\times 20$
$\Rightarrow m_1^{\prime }=1\text{kg}$
So, final mass of water = Initial mass of water + Mass of ice that melts $=5+1=6\text{kg}$