$2 k g$ of ice at $20^{\circ} C$ is mixed with $5 k g$ of water at $20^{\circ}$ in an insulating vessel having 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 $1 k c a l / k g /^{\circ} C$ and $0.5 k c a l / k g /^{\circ} C$ respectively, while the latent heat of fusion of ice is $80 k c a l / k g$ .

7 k g

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6 k g

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4 k g

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2 k g

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Solution

The correct option is B $6 k g$
Heat required to convert $5 k g$ of water at $20^{\circ} C$ to $5 k g$ of water at $0^{\circ} C$
$= m C_{w} △ T = 5 \times 1 \times 20 = 100 k c a l$ .
Heat released by $2 k g$ ice at $- 20^{\circ} C$ to convert $2 k g$ of ice at $0^{\circ} C$
$= m C_{i c e} △ T = 2 \times 0.5 \times 20 = 20 k c a l$ .
How much ice at $0^{\circ} C$ will convert into water at $0^{\circ} C$ for giving another $80 k c a l$ of heat $Q = m L$
$\Rightarrow 80 = m \times 80 \Rightarrow m = 1 k g$
Therefore, the amount of water at $0^{\circ} C - 5 k g + 1 k g = 6 k g$ . Thus, at equilibrium we have
$(6 k g$ water at $0^{\circ} C + 1 k g$ ice at $0^{\circ} C)$ .

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