$1\,kg$ of water at $100\, ^{\circ}C$ is converted into steam at $100^{\circ}\,C$ 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 $........kJ$ (Given latent heat of vaporisaiton $=2257\,kJ / kg$. Atmospheric pressure $=1 \times 10^5\,Pa$ )
JEE MAIN 2023, Medium
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$\Delta Q =\Delta U +\Delta W$
$\therefore \Delta U =\Delta Q -\Delta W$
$= mL _{ V }- P \Delta V$
$=(1\,Kg )\left(2257 \times 10^3\,J / kg \right)$
$-\left(1 \times 10^5\,Pa \right)\left(1.671\,m ^3-1 \times 10^{-3}\,m ^3\right)$
$=2257 \times 10^3\,J -167 \times 10^3\,J$
$=2090\,KJ$
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