A 44-g block of ice at -15°C is dropped into a calorimeter (of negligible heat capacity) containing 100 g of water at 5°C. When equilibrium is reached, how much of the ice will have melted? The specific heat of ice is 2090 J/(kg K) and the latent heat of fusion of water is
33) 5 × 10⁴ J/kg.

Question

Quizplus · Accepted Answer

Explanation:The heat required to melt the ice is equal to the heat lost by the water.Q = mL = (44 g)(335 × 10^4 J/kg) = 1.47 × 10^7 JQ = mcΔT = (100 g)(4187 J/(kg K))(5°C - 0°C) = 2.09 × 10^6 JTherefore, the mass of ice that melted is:m = Q/L = (1.47 × 10^7 J)/(335 × 10^4 J/kg) = 44 gSo, the correct answer is A.

A 44-G Block of Ice at -15°C Is Dropped into a Calorimeter