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A steel container of water equivalent 10 g holds 20 g of ice at -30 deg C. Then 30 g of water at 80 deg C is poured into the container. Find the final equilibrium temperature. (Given: S_ice = 0.5 cal/(g*deg C), S_water = 1 cal/(g*deg C), L_ice = 80 cal/g)

1. 0 deg C
2. -5.55 deg C
3. 3.33 deg C
4. 5.55 deg C

Correct answer: 3.33 deg C

Solution

Hot water releases 2400 cal cooling to 0 deg C. After warming ice and container (600 cal) and melting all ice (1600 cal), 200 cal remains. This raises 60 g of water+container from 0 deg C by 200/60 = 3.33 deg C.

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