Question

72.3 g of ice at -15.0oC has heat energy added to it until it becomes steam at 145oC. Calculate the total amount of heat energy needed (in Joules) to accomplish this.

Answer 1

Answer:

• 226 kJ

Explanation:

We will need these constants for water:

• Specific heat capacity of ice = 2.108 kJ/kg•ºC

• Specific heat capacity of water = 4.186 kJ/kg•ºC

• Specific heat capacity of steam = 1.996 J/kg•ºC

• Latent heat of melting, ΔHf = 334 kJ/kg

• Latent heat of evaporation, ΔHvap = 2256 kJ/kg

Calculations:

• 72.3g = 0.0723kg

1. Heating the ice from -15ºC to 0ºC

• Q₁ = m×C×ΔT = 0.0723 kg × 2.108 kJ/kg.ºC  × 15ºC = 2.286kJ

2. Melting the ice at 0ºC

• L₂ = m × ΔHf = 0.0723g × 334 kJ/kg = 24.148kJ

3. Heating the liquid water from 0ºC to 100ºC

• Q₃ = 0.0723g × 4.186 J/kg.ºC × 100ºC = 30.265kJ

4. Evaporating the liquid at 100ºC

• L₄ = 72.3g × 2256 KJ/g = 163.109kJ

5. Heating the steam from 100ºC to 145ºC

• Q₅ = 0.0723g × 1.996 × 45ºC = 6.494kJ

Total = 2.286kJ + 24.148kJ + 30.265kJ + 163.109 kJ + 6.494 kJ = 226.302 kJ = 226 kJ