Question

10 g of ice at -20 degree c is mixed with 2gof steam at 100 degree c. find d temperature of mixture

Answer 1

The ice is first heated to a temperature of t₀ = 0 °C, then fuses and possibly further heats up. The steam condenses into water at 100 °C, and then possibly to be cooled down. Depending on the amount of heat of the steam ice not fuses or fuses partially.
Ice:
m₁ = 10 g = 0.01 kg
t₁ = -20 °C
C₁ = 2.1 kJ / (kg · °C)  - heat capacity of ice
C_f = 334 kJ / kg  - heat of fusion of ice
c₁ = 4,2 kJ / (kg · °C) - heat capacity of water
Steam:
m₂ = 2 g = 0.002 kg
t₂ = 100 °C
C_c = 2260 kJ / kg  - heat of condesation (= heat of vaporization)
c₂ = c₁ = 4,2 kJ / (kg · °C) - heat capacity of water

The heat cast by water vapour, which condensed:
H₂₁ = m₂C_c = 0.002 · 2260 = 4,52 kJ
The heat collected by the ice, which is warm and fused:
H₁₁ = m₁C₁(t₀ - t₁) = 0.01 · 2.1 (0 + 20) = 0.42 kJ
The heat collected by the ice, which is fused:
H₁₂ = m₁C_f = 0.01 · 334 = 3.34 kJ
H₂₁ > H₁₁ + H₁₂ 
and so all ice is fused and heat up to tₓ

The heat collected by the water from t₀ to tₓ:
H₁₃ = m₁c₁(tₓ - t₀) 

The heat cast by the water from t₂ to tₓ:
H₂₂ = m₂c₂(t₂ - tₓ)

On the basis of the laws of thermodynamic:
H₁₁ + H₁₂ + H₁₃ = H₂₁ + H₂₂
m₁C₁(t₀ - t₁) + m₁C_f + m₁c₁(tₓ - t₀) = m₂C_c + m₂c₂(t₂ - tₓ)
m₁c₁tₓ + m₂c₂tₓ = m₂C_c - m₁C₁(t₀ - t₁) - m₁C_f + m₁c₁t₀ + m₂c₂t₂
tₓ = {m₂(C_c + c₂t₂) - m₁[C₁(t₀ - t₁) + C_f - c₁t₀]} ÷ (m₁c₁ + m₂c₂)
tₓ = {0.002(2260 + 4,2 · 100) - 0.01[2.1(0 + 20) + 334 - 0]} ÷ (0.01 · 4.2 + 0.002 · 4.2)
tₓ ≈ 31.7 °C  [kg·kJ/kg ÷ (kg ·kJ/(kg·°C) = °C]

Answer:
The ending temperature of mixture is about 31.7 °C