A sample of 0.1 g of water at 100°C and normal pressure (1.013 x 10 Nm ?) requires 54 cal of heat energy to convert to steam at 100°C. If the
volume of the steam produced is 167.1 cc, then change in internal energy of the sample is :-
(A) 104.3 J
(B) 208.7 J
(C) 42.2 J
(D) 84.5 J
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Answer:
(B) 208.7J
Explanation:
From first law of thermodynamics,
δQ=dU+δW
54×4.184=dU+PdV
54×4.184=dU+1.013×10 ^5 ×167.1×10 ^−6 0
On solving, we will get dU=208.7J
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The Answers is B 208.7
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