1 kg of water at is converted into steam at by boiling at atmospheric pressure. The volume of water changes from as a liquid to as steam. [2023]

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Solution

Q.
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 $1.00 \times 10^{- 3} m^{3}$ 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 vaporisation $=2257 kJ/kg$ , atmospheric pressure $= {1×10}^{5} Pa$ ) [2023]

1 + 2090 kJ

2 - 2090 kJ

3 - 2426 kJ

4 + 2476 kJ

Ans.
(1)

$Δ Q = Δ U + Δ W$

$∴$ $Δ U = Δ Q - Δ W = m L_{v} - P Δ V$

$= {(1 kg)(2257×10}^{3} J/kg)$

$- (1 \times 10^{5} Pa) ({1.671 m}^{3} - 1 \times 10^{- 3} m^{3})$

$= 2257 \times 10^{3} J - 167 \times 10^{3} J$ $= 2090 kJ$

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