$540$ calories of heat convert $1 $ cubic centimeter of water at ${100^o}C$ into $1671 $ cubic centimeter of steam at ${100^o}C$ at a pressure of one atmosphere. Then the work done against the atmospheric pressure is nearly ...... $cal$
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(b) Amount of heat given $ = 540\;calories$
Change in volume $\Delta V = 1670\;c.c$
Atmospheric pressure $P = 1.01 \times {10^6}\;dyne/c{m^2}$
Work done against atmospheric pressure
$W = P\Delta V$$ = \frac{{1.01 \times {{10}^6} \times 1670}}{{4.2 \times {{10}^7}}} \approx 40\;cal$
Change in volume $\Delta V = 1670\;c.c$
Atmospheric pressure $P = 1.01 \times {10^6}\;dyne/c{m^2}$
Work done against atmospheric pressure
$W = P\Delta V$$ = \frac{{1.01 \times {{10}^6} \times 1670}}{{4.2 \times {{10}^7}}} \approx 40\;cal$
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