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Work done in converting 1 g of ice at $- 10^{0} C$ into steam at $100^{0} C$ is

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721 J

161 J

3045 J

6056 J

answer is A.

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Detailed Solution

$W = Q_{1} + Q_{2} + Q_{3} + Q_{4}$

$= m S_{i c e} (0 - (- 10)) + m L_{i c e} + m S_{W} (100 - 0) + m L_{v a p o u r}$

On solving with values as

$S_{i c e} = 2100 J / k g /^{0} C, S_{w a t e r} = 4200 J / k g /^{0} C$

$L_{i c e} = 336 K J / K g, L_{v a p o u r} = 2260 K J / K g$

we get

$W = 3045 J$

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