A 26.6 g sample of aluminum at 100.4 °C is added to 100.6 g of water at 21.5 °C in a constant pressure calorimeter. What is the final temperature of the water in °C? The specific heat capacity of aluminum is 0.903 J/goC.

Given:

m₁ = 26.6 g, the mass of aluminum

T₁ = 100.4 °C, the temperature of the aluminum

c₁ = 0.903 J / (g-°C), the specific heat of aluminum

m₂ = 100.6 g, the mass of water

T₂ = 21.5 °C, the temperature of water

The specific heat of water is c₂ = 4.184 J / (g-C)

Let T = the final temperature of the water and aluminum.

Calculate heat loss of the aluminum.

Q₁ = (26.6 g) * (0.903 J / (g-C)) * (100.4-T C) = 24.1098 (100.4 - T) J

Calculate heat gained by the water

Q₂ = (100.6 g) * (4.184 J / (g-C)) * (T - 21.5 C) = 420.1904 (T - 21.5) J

Conservation of energy requires that Q₁ = Q₂.

Therefore

420.1904 (T - 21.5) = 24.1098 (100.4 - T)

444.3T = 1.1455 x 10⁴

T = 25.782 °C

Answer: 25.8 °C