1 kg of water at 100 is converted into steam at 100 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 vaporisation = 2257 kJ/kg, Atmospheric pressure = 1 × 10⁵ Pa) (a) + 2090 kJ (b) - 2426 kJ (c) + 2476 kJ (d) - 2090 kJ

Answer: (a). Heat absorbed: Q = mL = 1 × 2257 = 2257 kJ Work done: W = PΔ V = 10⁵ × (1.671 - 0.001) = 1.67×10⁵ J = 167 kJ Δ U = Q - W = 2257 - 167 = +2090 kJ So the answer is (a) +2090 kJ.

1 kg of water at 100 is converted into steam at 100 by boiling at… | JEE Main 2023 (11 Apr, Shift 1)

Q11JEE Main 2023 · 11 APR · SHIFT 1Medium

Heat, Work & First Law of Thermodynamics

$1\text{ kg}$ of water at $100^\circ\text{C}$ is converted into steam at $100^\circ\text{C}$ by boiling at atmospheric pressure. The volume of water changes from $1.00 \times 10^{-3}\text{ m}^{3}$ as a liquid to $1.671\text{ m}^{3}$ as steam. The change in internal energy of the system during the process will be

(Given latent heat of vaporisation $= 2257\text{ kJ/kg}$ , Atmospheric pressure = $1 \times 10^{5}\text{ Pa})$

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Answer: (a)

Heat absorbed: $Q = mL = 1 \times 2257 = 2257\text{ kJ}$
Work done: $W = P\Delta V = 10^{5} \times (1.671 - 0.001) = 1.67\times10^{5}\text{ J} = 167\text{ kJ}$
$\Delta U = Q - W = 2257 - 167 = +2090\text{ kJ}$
So the answer is (a) +2090 kJ.

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