Calorimetry: A block of ice at 0.000°C is added to a well-insulated 147-g aluminum calorimeter cup that holds 200 g of water at 10.0°C. The water and aluminum cup are in thermal equilibrium, and the specific heat of aluminum is 910 J/(kg ∙ K). If all but 2.00 g of ice melt, what was the original mass of the block of ice? The specific heat of water is 4190 J/(kg ∙ K), its latent heat of fusion is 334 kJ/kg, and its latent heat of vaporization is 2260 kJ/kg.
A) 31.1 g
B) 35.6 g
C) 38.8 g
D) 42.0 g
E) 47.6 g

Question

Quizplus · Accepted Answer

The correct answer is A because the heat lost by the ice is equal to the heat gained by the water and the aluminum cup. The heat lost by the ice is equal to the mass of the ice that melted multiplied by the latent heat of fusion of ice. The heat gained by the water is equal to the mass of the water multiplied by the specific heat of water multiplied by the change in temperature of the water. The heat gained by the aluminum cup is equal to the mass of the aluminum cup multiplied by the specific heat of aluminum multiplied by the change in temperature of the aluminum cup.

Calorimetry: a Block of Ice at 0