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  4. 1 g of water (volume 1 cm 3 ) becomes 1671 cm 3 of steam when boiled at a pressure of 1 atm. The latent heat of vaporization is 540 cal / g, then the external work done is: (1 atm =1.013 × 105 N / m 2)

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Q. $1\, g$ of water (volume $1\, cm ^{3}$ ) becomes $1671\, cm ^{3}$ of steam when boiled at a pressure of $1 \, atm$. The latent heat of vaporization is $540\, cal / g$, then the external work done is:
$\left(1 atm =1.013 \times 10^{5} \, N / m ^{2}\right)$

BITSATBITSAT 2011

Solution:

Work done, $W = p \Delta V$
$ = 1.013 \times 10^5 \times (1671 - 1)$
$ = 1.013 \times 10^5 \times 1670 \times 10^{-6} $
$ = 169.2 \, J $