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$1 \mathrm{~g}$ of water at $100^{\circ} \mathrm{C}$ is completely converted into steam at $100^{\circ} \mathrm{C} .1 \mathrm{~g}$ of steam occupies a volume of 1650 cc. (Neglect the volume of $1 \mathrm{~g}$ of water at $100^{\circ} \mathrm{C}$ ). At the pressure of $10^5 \mathrm{~N} / \mathrm{m}^2$, latent heat of steam is 540 cals $/ \mathrm{g}$ ( 1 calorie $=4.2$ joule). The increase in the internal energy (in joule) is
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Verified Answer
The correct answer is:
2203
Heat required to convert $1 \mathrm{~g}$ of water at $100^{\circ} \mathrm{C}$ completely into steam at $100^{\circ} \mathrm{C}$
$$
\begin{aligned}
& d Q=m L \\
& =1 \times 540 \text { calorie } \\
& =540 \times 4.2=2268 \mathrm{~J}
\end{aligned}
$$
Work done in expanding volume
$$
\begin{aligned}
d W & =p d V=10^5\left(V_2-V_1\right) \\
& =10^5(1650-1000) \times 10^6 \\
& =650 \times 10^{-1}=65
\end{aligned}
$$
Increase in internal energy
$$
d U=d Q-d W=2268-65=2203 \mathrm{~J}
$$
$$
\begin{aligned}
& d Q=m L \\
& =1 \times 540 \text { calorie } \\
& =540 \times 4.2=2268 \mathrm{~J}
\end{aligned}
$$
Work done in expanding volume
$$
\begin{aligned}
d W & =p d V=10^5\left(V_2-V_1\right) \\
& =10^5(1650-1000) \times 10^6 \\
& =650 \times 10^{-1}=65
\end{aligned}
$$
Increase in internal energy
$$
d U=d Q-d W=2268-65=2203 \mathrm{~J}
$$
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