Question

You place 740. g of ice at its melting point in a thermally insulated container. What is the mass of steam at 100°C that would need to be mixed with the ice to produce liquid water at 30.0°C? Any constants you may need can be found in the textbook.

Answer 1

Mass of ice = m₁ = 740 g = 0.74 kg

Mass of steam = m₂

Initial temperature of ice = T₁ = 0 ^oC

Initial temperature of steam = T₂ = 100 ^oC

Final temperature of the mixture = T₃ = 30 ^oC

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

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

Latent heat of vaporization of steam = L_s = 2260000 J/kg

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

m₁L_i + m₁C(T₃ - T₁) = m₂L_s + m₂C(T₂ - T₃)

(0.74)(334000) + (0.74)(4186)(30 - 0) = m₂(2260000 + 4186(100 - 30))

340089.2 = 2553020m₂

m₂ = 0.1332 kg

m₂ = 133.2 g

Mass of steam needed = 133.2 g