An insulated container is filled with a mix of 400 g of water at 20.0°C and 60 g of ice at 0.00°C. Assuming negligible heat is exchanged with the container, what is the temperature of the mixture when it reaches thermal equilibrium?

Question

Quizplus · Accepted Answer

The heat required to melt the ice is equal to the heat lost by the water. The heat capacity of water is 4.18 J/g°C, and the heat of fusion of ice is 334 J/g. Therefore, the heat required to melt the 60 g of ice is:Q = mL = (60 g)(334 J/g) = 20040 JThe heat lost by the 400 g of water is:Q = mcΔT = (400 g)(4.18 J/g°C)(ΔT)Setting these two quantities equal to each other, we get:20040 J = (400 g)(4.18 J/g°C)(ΔT)ΔT = 12.0°CTherefore, the temperature of the mixture when it reaches thermal equilibrium is 20.0°C + 12.0°C = 32.0°C.

An Insulated Container Is Filled with a Mix of 400