1 cc of water becomes 1681 cc of steam when boiled at pressure of $10^{5} \mathrm{~N} \mathrm{~m}^{-

Question

1 cc of water becomes 1681 cc of steam when boiled at pressure of $10^{5} \mathrm{~N} \mathrm{~m}^{-2}$. The increasing internal energy of the system is\ (L.T of steam is $540 \mathrm{cal} / \mathrm{g}, 1$ calorie $=4.2 \mathrm{~J}$ )\

Professor PC Thomas Classes · Answer

Work done by the water\ $\Delta \mathrm{W}-\mathrm{P}\left(\mathrm{V}_{\mathrm{g}}-\mathrm{V}_{\ell}ight)$\ $=10^{5}(1681-1) 	imes 10^{-6} \mathrm{~J}=1680 	imes 10^{-1} \mathrm{~J}=168 \mathrm{~J}=40 \mathrm{cal}$ Hea\ t supplied to the 1 g of water,\ $\Delta \mathrm{Q}=\mathrm{mL}=(1 \mathrm{~g})\left(540 \mathrm{calg}^{-1}ight)=540 \mathrm{cal}$\ According o first law of thermodynamics $\Delta Q=\Delta W+\Delta U$\ $	herefore$ Increase in internal energy\ $\Delta \mathrm{U}=\Delta \mathrm{Q}-\Delta \mathrm{W}=540 \mathrm{cal}-40 \mathrm{cal}=500 \mathrm{cal}$\