An aluminium vessel of mass 0.5 kg holds 0.2 kg of water at 20^ C . A block of iron of mass 0.2 kg at 100^ C is gently placed into the water. Determine the equilibrium temperature of the mixture. The specific heat capacities of aluminium, iron, and water are 910 J kg ⁻¹ K ⁻¹ , 470 J kg ⁻¹ K ⁻¹ , and 4200 J kg ⁻¹ K ⁻¹ , respectively.

Step-by-step solution

Using the principle of calorimetry, heat lost by the hot body is equal to the heat gained by the cold bodies. Let the equilibrium temperature of the mixture be T . Heat lost by the iron block: Q_ Fe = m_ Fe c_ Fe (100 - T) Q_ Fe = 0.2 470 (100 - T) = 94 (100 - T) Heat gained by the aluminium vessel and water: Q_ Al + Q_ w = m_ Al c_ Al (T - 20) + m_ w c_ w (T - 20) Q_ Al + Q_ w = (0.5 910 + 0.2 4200) (T - 20) Q_ Al + Q_ w = (455 + 840) (T - 20) = 1295 (T - 20) Equating heat lost to heat gained: 94 (100 - T) = 1295