10 gm of ice at $0 ° C$ is kept in a calorimeter of water equivalent 10 gm. How much heat should be supplied to the apparatus to evaporate the water thus formed?
(Neglect loss of heat, $L_{V} = 540 cal / gm, and L_{f} = 80 cal / gm$ )

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$6200 c a l$

$13600 c a l$

$8200 c a l$

$7200 c a l$

answer is D.

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Detailed Solution

Heat supplied is $L_{f u s i o n} + M c Δ T + M L_{v a p}$
$Q_{1} = 10 \times 336 + 10 \times 4.2 \times 100 + 10 \times 2260$
$Q_{1} = 30160 J o r 7200 c a l$
Heat for calorimeter $Q_{2} = 10 \times 1 \times 100 = 1000 c a l$

$Q = Q_{1} + Q_{2} = 8200 c a l$

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