A sample of 0.1 g of water at and normal pressure requires 54 cal of heat energy to convert to steam at . If the volume of the steam produced is , the change in internal energy of the sample is [2018]

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Solution

Q.
A sample of 0.1 g of water at $100 ° C$ and normal pressure $(1.013 \times 10^{5} {N m}^{- 2})$ requires 54 cal of heat energy to convert to steam at $100 ° C$ . If the volume of the steam produced is $167.1$ $cc$ , the change in internal energy of the sample is [2018]

1 104.3 J

2 208.7 J

3 42.2 J

4 84.5 J

Ans.
(2)

Using first law of thermodynamics, $Δ Q = Δ U + Δ W$

$\Rightarrow 54 \times 4.18 = Δ U + 1.013 \times 10^{5} (167.1 \times 10^{- 6} - 0)$

$\Rightarrow Δ U = 208.7$ $J$

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