Question

1gm water at ${100}^{o}C$ is heated to convert it into steam at ${100}^{o}C$ and 1 atm. Find the change in internal energy of water. It is given that volume of 1gm water at ${100}^{o}C$ is $1cc$, volume of 1gm steam at ${100}^{o}C=1671cc$. Latent heat of vaporization $=540cal/g$. (Mechanical equivalent of heat $J=4.2J/cal.$)

Answer 1

From first law of thermodynamic $\Delta Q=\Delta U+W$
$\Delta Q=mL=1\times 540cal.=540cal.$
$W=P\Delta V=\frac{{10}^6(1671-1)\times {10}^{-6}}{4.2}=\frac{{10}^5\times \left(1670\right)\times {10}^{-6}}{4.2}\approx 40cal$.
$\Delta U=540-40=500cal$.