Trying to beat the heat of the last summer, a physics grad student went to the local toy store and purchased a plastic child's 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 30g, into the pool. (The ice cubes were originally at 0C.) He continued to add ice cubes, until the temperature stabilized to 16C. 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.)

Question

Harley Meyer

Physics

30 December, 14:28

Trying to beat the heat of the last summer, a physics grad student went to the local toy store and purchased a plastic child's 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 30g, into the pool. (The ice cubes were originally at 0C.) He continued to add ice cubes, until the temperature stabilized to 16C. 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.)

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Answers (1)

Know the Answer?

Ashton Galvan · Answer 1

625 piece.

Explanation:

Let n be the required no of piece

200 litre = 200 x 10⁻³ m³ = 200 x 10⁻³ x 10³ kg = 200 Kg

mass of ice piece = n x 30 x / 1000

Heat lost by ice pieces = (n x 30) / 1000 x (80 + 1 x 16⁰C)

Heat gained by water = 200 x 1 x 9⁰C

Heat lost = Heat gained

(n x 30) / 1000 x (80 + 16) = 200 x 9

2.88 n = 1800

n = 625