A 2.0-kg mass of iron specific heat = 0.12 kcal/kg°C) at a temperature of 430°C is dropped into 0.4 kg of ice and 0.4 kg of water both at 0°C. With no heat losses to the surroundings, the equilibrium temperature of the mixture is approximately

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Quizplus · Accepted Answer

The heat lost by the iron is equal to the heat gained by the ice and water. The heat lost by the iron is:Q = mcΔt = (2.0 kg)(0.12 kcal/kg°C)(430°C - 0°C) = 103.2 kcalThe heat gained by the ice is:Q = mL = (0.4 kg)(80 kcal/kg) = 32 kcalThe heat gained by the water is:Q = mcΔt = (0.4 kg)(1.0 kcal/kg°C)(0°C - 69°C) = -27.6 kcalThe total heat gained by the ice and water is:Q = 32 kcal + (-27.6 kcal) = 4.4 kcalThe equilibrium temperature of the mixture is the temperature at which the heat lost by the iron is equal to the heat gained by the ice and water. This temperature is:T = Q/(mc) = 4.4 kcal/[(2.0 kg)(0.12 kcal/kg°C)] = 69°C

A 20-Kg Mass of Iron Specific Heat = 0