Ice at $0^{°} C$ is added to $200 g$ of water initially at $70^{°} C$ in a vacuum flask. When $50 g$ of ice has been added and has all melted the temperature of the flask and contents is $40^{°} C.$ When a further $80 g$ of ice has been added and has all melted the temperature of the whole becomes $10^{°} C.$ Find the latent heat of fusion of ice. (in cal/g)

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answer is 90.

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Detailed Solution

Let heat capacity of flask be C and latent heat of fusion of ice L.

Then, $C (70 - 40) + 200 \times 1 \times (70 - 40) = 50 L + 50 \times 1 \times (40 - 0) o r 3 C - 5 L = - 400 . . . . (i)$

Further, $C (40 - 10) + 250 \times 1 \times (40 - 10) = 80 L + 80 \times 1 \times (10 - 0) o r 3 C - 8 L = - 670 . . . . (i i)$

Solving Eq. (i) and (ii), we have, $L = 90 cal / g$

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