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A sample of 0.1 g of water at $100^\circ C$ and normal pressure ($1.013 \times 10^{5}\ N m^{-2}$) requires 54 cal of heat energy to convert to steam at $100^\circ C$. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is
Explanation
$\Delta U = Q - P\Delta V$.
Detailed Solution
$\Delta Q = \Delta U + \Delta W$
$54\times4.18 = \Delta U + 1.013\times10^{5}(167.1\times10^{-6} - 0)$
$\Delta U = 225.7 - 16.9 = 208.7$ J
