Question

A 95-g aluminum calorimeter is filled with ice taken from the freezer at -8.5⁰C. The calorimeter is then filled with 100 g of steam at a temperature of 103⁰C. Assume the aluminum calorimeter has the same initial temperature as the ice. The final situation is observed to be all water at 90.0⁰C.

1. What was the mass of the ice?

Answer 1

Mass of the aluminium calorimeter = m₁ = 95 g = 0.095 kg

Mass of the ice = m₂

Mass of the steam = m₃ = 100 g = 0.1 kg

Initial temperature of ice and calorimeter = T₁ = -8.5 ^oC

Initial temperature of steam = T₂ = 103 ^oC

Final temperature = T₃ = 90 ^oC

Melting point of ice = T₄ = 0 ^oC

Boiling point of water = T₅ = 100 ^oC

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

Specific heat of ice = C_i = 2090 J/(kg.^oC)

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

Specific heat of steam = C_s = 2010 J/(kg.^oC)

Latent heat of fusion of ice = L_i = 334000 J/kg

Latent heat of vaporization of water = L_w = 2260000 J/kg

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

m₁C_a(T₃ - T₁) + m₂C_i(T₄ - T₁) + m₂L_i + m₂C_w(T₃ - T₄) = m₃C_s(T₂ - T₅) + m₃L_w + m₃C_w(T₅ - T₃)

(0.095)(900)(90 - (-8.5)) + m₂(2090)(0 - (-8.5)) + m₂(334000) + m₂(4186)(90 - 0) = (0.1)(2010)(103 - 100) + (0.1)(2260000) + (0.1)(4186)(100 - 90)

8421.75 + 728505m₂ = 230789

m₂ = 0.305 kg

Mass of the ice = 0.305 kg