Question

An aluminum calorimeter cup has a mass of 40 grams and contains 55 grams of ice originally at -5 degrees C. An unknown solid originally at 100 degrees C has a mass of 390 grams is placed in the calorimeter cup with ice. The final temperature of the mixture is 20 degrees C. Calculate the specific heat of the unknown substance.

Answer 1

Mass of the aluminium calorimeter cup = m₁ = 40 g = 0.04 kg

Mass of the ice = m₂ = 55 g = 0.055 kg

Mass of the unknown solid = m₃ = 390 g = 0.39 kg

Initial temperature of the calorimeter and the ice = T₁ = -5 ^oC

Initial temperature of the unknown solid = T₂ = 100 ^oC

Final equilibrium temperature = T₃ = 20 ^oC

Melting point of ice = T₄ = 0 ^oC

Specific heat of aluminium = C_a = 900 J/(kg.^oC)

Specific heat of ice = C_i = 2108 J/(kg.^oC)

Specific heat of water = C_w = 4186 J/(kg.^oC)

Specific heat of the unknown solid = C

Latent heat of fusion of water = L = 334000 J/kg

The heat gained by the calorimeter and the ice is equal to the heat lost by the unknown solid.

m₁C_a(T₃ - T₁) + m₂C_i(T₄ - T₁) + m₂L + m₂C_w(T₃ - T₄) = m₃C(T₂ - T₃)

(0.04)(900)(20 - (-5)) + (0.055)(2108)(0 - (-5)) + (0.055)(334000) + (0.055)(4186)(20 - 0) = (0.39)C(100 - 20)

24454.3 = 31.2C

C = 783.79 J/(kg.^oC)

The specific heat of the unknown substance = 783.79 J/(kg.^oC)