The latent heat of vaporization of water is $2240 \,J/gm$. If the work done in the process of vaporization of $1\, gm$ is $168\, J$, then increase in internal energy is  .... $J$
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$\mathrm{L}=2240 \mathrm{J}, \mathrm{m}=1 \mathrm{gm}$

$\mathrm{d} \mathrm{W}=168 \mathrm{J}$

$\mathrm{dQ}=\mathrm{mL}=\mathrm{dU}+\mathrm{d} \mathrm{W}$

or $1 \times 2240=d U+168$

$\mathrm{dU}=2072 \mathrm{J}$

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