Question

Ice of water at ${100}^o$ is given 540 cal of beat and the steam formed occupation 1671 cc at the atmosphere pressure. Then workdone against atmosphere pressure in this procces is nearly

• A. 540 cal
• B. 500 cal
• C. 40 cal
• D. 100 cal

Answer 1

The correct option is C 40 cal

$\text{Change in vol.}=(1671-1)cc$

$=1670\times {10}^{-6}{m}^3$

$\text{Atmospheric pressure:-}{10}^{5}pa\text{.}$

$\text{Net work done}=P\Delta V$

$={10}^5\times 1670\times {10}^{-6}$

$=167J$

$=39.78cal[1cal=4.2J]$

Answer $1-\left(C\right)$ $40cal$