1 kg of water at 100° C is converted into steam at 100° C by boiling at atmospheric pressure. The volume of water changes from 1.00 × 10⁻³ m³ as a liquid to 1.671 m³ as steam. The change in internal energy of the system during the process will be
(Given latent heat of vaporisaiton =2257 kJ / kg, Atmospheric pressure = 1 × 10⁵ Pa)

Solution

Physics sir · 3 steps

Boiling water takes in latent heat, and part of it is spent pushing back the air. The rest stays inside as ΔU.

1. Heat taken in. The water absorbs its latent heat of vaporisation, so Q = mL = 1 × 2257 = +2257 kJ. It is positive because heat flows into the system.
2. Work done by steam. The steam expands against a constant atmospheric pressure, so W = PΔ V = 1 × 10⁵ × (1.671 - 0.001) = 1.67 × 10⁵ J = 167 kJ. It is positive because the system does work on the surroundings.

   Δ U = Q - W = 2257 kJ - 167 kJ = 2090 kJ

3. So. The internal energy rises by 2090 kJ, which is option (a). The temperature stays at 100°C, but this energy goes into pulling the molecules apart, so it is stored as potential energy.

The idea to keep. In a phase change at constant pressure, Q = mL and W = PΔ V. Then Δ U = Q - W. A constant temperature does not mean ΔU is zero once the phase changes.

Answer from the NTA final answer key. Solution by PrepWiser.