330 grams of boiling water (temperature 100°C, specific heat capacity 4.2 J/K/gram) are poured into an aluminum pan whose mass is 840 grams and initial temperature 29°C (the specific heat capacity of aluminum is 0.9 J/K/gram). After a short time, what is the temperature of the water?

T = 74°C

Explanation:

Given Mw = mass of water = 330g, Ma = mass of aluminium = 840g

Cw = 4.2gJ/g°C = specific heat capacity of water and Ca = 0.9J/g°C = specific heat capacity of aluminium

Initial temperature of water = 100°C.

Initial temperature of aluminium = 29°C

When the boiling water is poured into the aluminum pan, heat is exchanged and after a short time the water and aluminum pan both come to thermal equilibrium at a common temperature T.

Heat lost by water equal to the heat gained by aluminium pan.

Mw * Cw * (100 - T) = Ma * Ca * (T-29)

330*4.2 * (100 - T) = 890*0.9 * (T-29)

1386 (100 - T) = 801 (T - 29)

1386/801 (100 - T) = T - 29

1.73 (100 - T) = T - 29

173 - 1.73T = T - 29

173+29 = T + 1.73T

202 = 2.73T

T = 202/2.73

T = 74°C