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A 10 g block of ice at 0 degC is placed in a calorimeter whose water equivalent is 10 g. How much heat must be supplied to the system to first melt the ice and then completely evaporate the resulting water? (Neglect heat losses; use latent heat of fusion 80 cal/g, latent heat of vaporization 540 cal/g, specific heat of water 1 cal/g/degC.)

1. 6200 cal
2. 7200 cal
3. 13600 cal
4. 8200 cal

Correct answer: 8200 cal

Solution

Q = 10*80 (melt) + (10+10)*1*100 (raise to 100 degC) + 10*540 (evaporate) = 800 + 2000 + 5400 = 8200 cal.

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