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

Question

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 - T) = 103.2 kcalThe heat gained by the ice is:Q = mLf = (0.4 kg)(80 kcal/kg) = 32 kcalThe heat gained by the water is:Q = mcΔT = (0.4 kg)(1 kcal/kg°C)(T - 0°C) = 0.4T kcalSetting the heat lost by the iron equal to the heat gained by the ice and water, we get:103.2 kcal = 32 kcal + 0.4T kcalSolving for T, we get:T = 173°CTherefore, the equilibrium temperature of the mixture is approximately 69°C.

A 20-Kg Mass of Iron (Specific Heat = 0 ∘\circ∘ C)at a Temperature of 430

A 20-Kg Mass of Iron (Specific Heat = 0 $\circ$ C)at a Temperature of 430