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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 vaporization = 2257 kJ/kg. Atmospheric pressure = $1 \times 10^{5} Pa$ )

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+2476 kJ

+2090 kJ

-2090 kJ

-2426 kJ

answer is C.

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Detailed Solution

$dQ = dU + PdV$

$\Rightarrow mL = dU + PdV$

$\Rightarrow dU = mL - PdV = 1 \times 2257 \times 10^{3} - 10^{5} \times 1670 \times 10^{- 3}$

$\Rightarrow dU = (2257 - 167) \times 10^{3} = 2090 kJ$

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