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OpenStudy (anonymous):

A 100-g ice cube at 0°C is placed in 650 g of water at 25°C. What is the final temperature of the mixture?

OpenStudy (anonymous):

heat gained by the ice and its melted water = heat lost by bath water (100gm)(79.7 cal/gm heat of fusion) + (100gm)(1cal/gm-oC)(T-0) = (650gm)(1cal/gm-oC)(25-T) We have used the masses, the specific heats and the post-prior temperatures. When I solved for T, I got 11oC, which seemed reasonable.

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