1 kg of water at $100^{\circ} \mathrm{C}$ is converted into steam at $100^{\circ} \mathrm{C}$ by boiling at atmospheric pressure. The volume of water changes from $1.00 \times 10^{-3} \mathrm{~m}^{3}$ as a liquid to $1.671 \mathrm{~m}^{3}$ as steam. The change in internal energy of the system during the process will be (Given latent heat of vaporisaiton $=2257 \mathrm{~kJ} / \mathrm{kg}$. Atmospheric pressure $=1 \times 10^{5} \mathrm{~Pa}$)

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Heat & Thermodynamics

1 kg of water at

10 0^{\circ} C

is converted into steam at

10 0^{\circ} C

by boiling at atmospheric pressure. The volume of water changes from

1.00 \times 1 0^{- 3} m^{3}

as a liquid to

1.671 m^{3}

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 \times 1 0^{5} Pa

)

Answer: (A)

By the first law,

Δ U = Δ Q - Δ W

, where

Δ Q = m L_{v}

is the heat supplied and

Δ W = P Δ V

is the work done against the constant atmospheric pressure. Here

Δ Q = (1) (2257 \times 1 0^{3}) = 2257 \times 1 0^{3}

J. The volume rises from

1.00 \times 1 0^{- 3}

^{3}

1.671

^{3}

, so

Δ W = (1 \times 1 0^{5}) (1.671 - 0.001) = 1.67 \times 1 0^{5}

= 167

kJ. Therefore

Δ U = 2257 - 167 = + 2090

kJ.

JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Single correct option