The amount of heat energy that will be extracted when 0.02kg of water vapor condenses to water at 97ºC is…

Analyze the question: Water vapour condenses to water at 97°C. We need the heat extracted. The question is ambiguous, so first work out what the data supports.

Concept: Heat released during condensation is Q = mL, where L is the specific latent heat of vaporization.

Calculation: Q = mL = 0.02 × 2.26 × 10⁶ = 45,200 J

That matches option B exactly.

Checking option A (45,452 J): 45,452 − 45,200 = 252 J. This is 0.02 × 4200 × 3 = 252 J, which is the heat released when the water cools by 3°C, from 100°C to 97°C.

So the full process is:

1. Vapour at 100°C condenses to water at 100°C: 45,200 J
2. Water cools from 100°C to 97°C: mcΔθ = 0.02 × 4200 × 3 = 252 J
3. Total = 45,200 + 252 = 45,452 J

Which reading is right? The question says the vapour condenses to water at 97°C. This means the final water is at 97°C, so the vapour (at its boiling point of 100°C) condenses and then the water cools down to 97°C. The cooling step must therefore be included, and the total is 45,452 J.

Answer: A (45,452 J)

Tip: When a question gives the specific heat capacity of water as well as the latent heat, check for a temperature change as well as a phase change. Steam condensing at 100°C and then cooling to a lower temperature needs both Q = mL and Q = mcΔθ. If the specific heat capacity is supplied, it is usually meant to be used.