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confused1

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Trying to beat the heat of summer, a physics grad student went to the local toy store and purchased a child's plastic swimming pool. Upon returning home, he filled it with 200 liters of water at 25 C. Realizing that the water would probably not be cool enough, he threw ice cubes from his refrigerator, each of mass 30 g, into the pool. (The ice cubes were originally at 0 C.) He continued to add ice cubes until the temperature stabilized at 16 C. He then got in the pool.

The density of water is 1000 kg/m3, the specific heat of water is 1.0 cal/g C, the specific heat of ice is 0.5 cal/g C, and the latent heat of fusion of water is 80 cal/g.

How many ice cubes did he add to the pool to get the temperature to 16 C? (Consider the pool and ice cubes an isolated system.)

HELP: Heat lost by water = heat gained by ice cubes. No heat is lost to the surroundings.

HELP: Since the water (subsystem 1) is at a higher temperature, heat will be lost to the ice cubes (subsystem 2). Calculate the heat H that the water gave up from 25 C to 16 C; calculate the heat h that each ice cube gained from 0 C to 16 C including melting. Then the number of ice cubes equals H/h.

The density of water is 1000 kg/m3, the specific heat of water is 1.0 cal/g C, the specific heat of ice is 0.5 cal/g C, and the latent heat of fusion of water is 80 cal/g.

How many ice cubes did he add to the pool to get the temperature to 16 C? (Consider the pool and ice cubes an isolated system.)

HELP: Heat lost by water = heat gained by ice cubes. No heat is lost to the surroundings.

HELP: Since the water (subsystem 1) is at a higher temperature, heat will be lost to the ice cubes (subsystem 2). Calculate the heat H that the water gave up from 25 C to 16 C; calculate the heat h that each ice cube gained from 0 C to 16 C including melting. Then the number of ice cubes equals H/h.

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