A person pours 330 g of water at 90°C into an 855-g aluminum container with an initial temperature of 10°C. The specific heat of aluminum is 900 J/(kg·K) and that of water is 4190 J/(kg·K). What is the final temperature of the system, assuming no heat is exchanged with the surroundings?

61.4 °C

Explanation:

Heat lost by the water = heat gained by the aluminum container.

cm(t₁-t₃) = c'm'(t₃-t₂) Equation 1

Where c = specific heat capacity of water, m = mass of water, t₁ = initial temperature of water, t₃ = final temperature of the mixture, c' = specific heat capacity of aluminum, m' = mass of aluminum container, t₂ = initial temperature of aluminum container

Given: c = 4190 J/kg.K, c' = 900 J/kg.K, m = 330 g = 0.33 kg, m' = 855 g = 0.855 kg, t₁ = 90 °C, T₂ = 10 °C.

Substitute into equation 1

4190×0.33(90-t₃) = 900×0.855(t₃-10)

1382.7(90-t₃) = 769.5(t₃-10)

124443-1382.7t₃ = 769.5t₃-7695

solving for t₃

769.5t₃+1382.7t₃ = 124443+7695

2152.2t₃ = 132138

t₃ = 132138/2152.2

t₃ = 61.4 °C

Hence the final temperature = 61.4 °C