Question

9. 100-g aluminum calorimeter contains 250-g of water. The two substances are in thermal equilibrium at 10° C. Two metallic blocks are placed in the water. One is a 50- piece of copper at 80° C. The other sample has a mass of 70-g and is originally at temperature of 100° C. The entire system stabilizes at a final temperature of 200 C. Determine the specific heat of the unknown second sample. (Assume the specific heat of aluminum, and copper are 0.215 cal/g °C, and 0.0924 cal/g°C, respectively, and that of water is I cal/g °C) (5 points) A

Answer 1

Mass of the aluminium calorimeter = m₁ = 100 g

Mass of the water = m₂ = 250 g

Mass of the piece of copper = m₃ = 50 g

Mass of the unknown sample = m₄ = 70 g

Initial temperature of the aluminium calorimeter and water = T₁ = 10 ^oC

Initial temperature of piece of copper = T₂ = 80 ^oC

Initial temperature of the unknown sample = T₃ = 100 ^oC

Final equilibrium temperature = T₄ = 20 ^oC

Specific heat of aluminium = C₁ = 0.0924 cal/(g.^oC)

Specific heat of water = C₂ = 0.215 cal/(g.^oC)

Specific heat of copper = C₃ = 1 cal/(g.^oC)

Specific heat of the unknown sample = C₄

The heat gained by the aluminium calorimeter and the water is equal to the heat lost by the copper and the unknown sample.

m₁C₁(T₄ - T₁) + m₂C₂(T₄ - T₁) = m₃C₃(T₂ - T₄) + m₄C₄(T₃ - T₄)

(100)(0.0924)(20 - 10) + (250)(1)(20 - 10) = (50)(0.215)(80 - 20) + (70)C₄(100 -20)

C₄ = 0.348 cal/(g.^oC)

Specific heat of the unknown sample = 0.348 cal/(g.^oC)