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course PHY 232
9/9
The thermal energy change of the calorimeter plus the water is .150 kg * 420 J/(kg C) * 56 C + .34 kg * 4187 J / (kg C) * 56 C = 83,250 J, approx.
The thermal energy change of the condensed water is -Hf * .035 kg + .035 kg * 4187 J / (kg C) * (-29 C) = -Hf * .035 kg - 2930 J, approx.
Net thermal energy change is zero, so we have
· 83,250 J - Hf * .035 kg - 4930 J = 0 which is easily solved to give us
· Hf = 79,000 J / (.035 kg) = 2,257,000 J / kg. **
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