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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 $100 \times 10^{- 3} m^{3}$ 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 vapourisation = $2257 k J / k g$ . Atmospheric pressure = $1 \times 10^{5}$ Pa)

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$+ 2476 k J$

$- 2090 k J$

$+ 2090 k J$

$- 2426 k J$

answer is C.

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Detailed Solution

$\begin{array}{l} d Q = d u + P d v \\ M L = d u + P d v \\ d u = M L - P d v \\ d u = 1 \times 22257 \times 10^{3} \\ - 10^{5} \times 1670 \times 10^{- 3} \\ d u = (2257 - 167) \times 10^{3} \\ d u = 2090 K J \end{array}$

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