2 kg of ice $a t - 20^{o} C$ is mixed with 5 kg of water at $20^{o} C$ in an insulating vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. lt is given that the specific heats of water and ice are 1 $k c a 1 / k g /^{\circ} C$ and 0.5 $k c a 1 / k g /^{\circ} C$ respectively while the latent heat of fusion of ice is 80 $k c a l / k g$ :

7 kg

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6 kg

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4 kg

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2 kg

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Solution

The correct option is A 6 kg
Energy required for increasing temperature of all the ice till $0^{o} C$ is $m s_{i c e} Δ T = 2 \times 0.5 \times 20 = 20 k c a l$
assume m mass of ice melts into water
Energy required for melting $= m L_{f u s i o n} = (80 m) k c a l$
all the energy is provided by the water
Energy supplied by water to cool it from $- 20^{o} C$ to $0^{o} C = m s_{w a t e r} Δ T = 5 \times 1 \times 20 = 100 k c a l$
hence $20 k c a l + (80 m) k c a l = 100 k c a l$
which gives $m = 1 k g$
hence total water in the mixture $= 5 + 1 = 6 k g$

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