$540$ calories of heat converts $1$ cubic centimeter of water at $100\,^oC$ into $1671$ cubic centimeter of steam at $100\,^oC$ at a pressure of one atmosphere. Then the work done against the atmospheric pressure is nearly .......... $cal$
Medium
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$W=P\left(V_{2}-V_{1}\right)$
$=10^{5}(1671-1) \times 10^{-6} \mathrm{J}$
$=\frac{10^{5} \times 1670 \times 10^{-6}}{4.2} \mathrm{cal}$
$W=\frac{1670 \times 10^{-1}}{4 \cdot 2}$
$=40 \mathrm{cal}$
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