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Q.

2 kg of ice at $- 20^{\circ} C$ is mixed with 5 kg of water at $20^{\circ} C$ in an insulting vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heat of water and ice are $1 k cal / kg /^{\circ} C$ and $0.5 k cal / kg /^{\circ} C$ while latent heat of fusion of ice is 80 kcal/kg :

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a

7 kg

b

6 kg

c

4 kg

d

2 kg

answer is B.

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Detailed Solution

Heat lost by water

$Q_{1} = 5 \times 10^{3} \times (20 - 0) = 10^{5} C a l$

Heat gained by ice during rise in temperature is

$Q_{2} = 2 \times 500 \times 20 = 0.2 \times 10^{5} c a l$

The remaining heat

$Q = Q_{1} - Q_{2} = 0.8 \times 10^{5} C a l$ will melt 1 kg of ice only.

Mass of water in mixture = 5 + 1 = 6 kg.

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2 kg of ice at –20∘C is mixed with 5 kg of water at 20∘C in an insulting vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heat of water and ice are 1 k cal/kg/∘C and 0.5 k cal/kg/∘C while latent heat of fusion of ice is 80 kcal/kg :