Question

$2\,kg$ ice at $-20^\circ C$ is mixed with $5\,kg$ water at $20^\circ C$ . Then final amount of water in the mixture would be; Given specific heat of ice $=0.5\,cal/g^\circ C$ , specific heat of water $=1\,cal/g^\circ C,$ Latent heat of fusion of ice $=80\,cal/g.$ (in $kg$ )

Answer 1

In this problem, ice will need energy to achieve the temperature of $0^\circ C$ and water will release energy to achieve the temperature of $0^\circ C$ .
Now, heat lost by $5\,kg$ water when its temperature decreases from $20^\circ C$ to $0^\circ C$ is given by
$Q_{1}=ms\Delta T=\left(5 \times \left(10\right)^{3}\right)\left(\right.1\left.\right)\left(\right.20\left.\right)$ $=10^{5}cal$
And, heat required by $2\,kg$ ice when its temperature increases from $-20^\circ C$ to $0^\circ C$
$Q_{2}=ms\Delta T=\left(2 \times \left(10\right)^{3}\right)\left(\right.0.5\left.\right)\left(\right.20\left.\right)$
$=0.20\times 10^{5}cal$
Now, the remaining energy $=Q=Q_{1}-Q_{2}=0.8\times 10^{5}cal$ will melt some ice. Mass of ice melted is given by using the expression
$Q=mL$
$\Rightarrow m=\frac{Q}{L}=\frac{0 . 8 \times 10^{5}}{80}=1000\,g=1\,kg$
Hence, the equilibrium temperature of the mixture will be $0^\circ C$ and mass of the water in equilibrium is is $5+1=6\,kg$ .