The increase in internal energy of 1 kg of water at 100° C when it is converted into steam at the same temperature and 1 atm $(100 kPa)$ will be : [The density of water and steam are $1000 kg / m^{3}$ and $0.6 kg / m^{3}$ respectively. The latent heat of vaporization of water is $2.25 \times 10^{6} J / kg$ .]

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$3.27 \times 10^{8} J$

$5 \times 10^{9} J$

$4 \times 10^{7} J$

$2.08 \times 10^{6} J$

answer is A.

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Detailed Solution

Given,

$\begin{array}{l} q = 2.25 \times 10^{6} \\ w = - 100 \times 10^{3} \times [\frac{1}{0.6} - \frac{1}{1000}] = - 1.67 \times 10^{5} \end{array}$

Thus, the value of internal energy,

$Δ U = q + w = 2.08 \times 10^{6} J$

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