A sample of 0.1 g of water at $100^{0} C$ and normal pressure $(1.013 \times 10^{5} N m^{- 2})$ requires 54 cal of heat energy to convert to steam at $100^{0} C$ . If the volume of the steam produced is 167.1 cc , the change in internal energy of the sample is

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104.3 J

208.7 J

42.2 J

84.5 J

answer is B.

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Detailed Solution

Using first law of thermodynamics,

$Δ Q = Δ U + Δ W$

$\begin{array}{l} \Rightarrow 54 \times 4.18 = Δ U + 1.013 \times 10^{5} (167.1 \times 10^{- 6} - 0) \\ \Rightarrow Δ U = 208.7 J \end{array}$

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