A block of ice at 0°C is added to a 150-g aluminum calorimeter cup that holds 200 g of water at 10°C. If all but 2.00 g of ice melt, what was the original mass of the block of ice?

Question

Quizplus · Accepted Answer

Explanation:The heat lost by the ice as it melts and warms to 10°C is equal to the heat gained by the water and the calorimeter cup:$$Q_{ice} = Q_{water} + Q_{cup}$$$$mL_f + mc_i(10°C - 0°C) = mc_w(10°C - 10°C) + mc_c(10°C - 10°C)$$$$mL_f = mc_w(10°C - 10°C) + mc_c(10°C - 10°C)$$$$mL_f = 0$$$$m = 0$$The heat lost by the remaining ice as it warms to 10°C is equal to the heat gained by the water and the calorimeter cup:$$Q_{ice} = Q_{water} + Q_{cup}$$$$mc_i(10°C - 0°C) = mc_w(10°C - 10°C) + mc_c(10°C - 10°C)$$$$mc_i(10°C) = 0$$$$m = 0$$The total heat lost by the ice is equal to the heat gained by the water and the calorimeter cup:$$Q_{ice} = Q_{water} + Q_{cup}$$$$mL_f + mc_i(10°C - 0°C) = mc_w(10°C - 10°C) + mc_c(10°C - 10°C)$$$$mL_f + mc_i(10°C) = 0$$$$mL_f + 2.00 g 	imes 2.09 J/g°C 	imes 10°C = 0$$$$mL_f = -41.8 J$$$$m = \frac{-41.8 J}{334 J/g} = 0.125 g$$The original mass of the block of ice was 0.125 g + 2.00 g = 2.125 g.

A Block of Ice at 0°C Is Added to a 150-G