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QNEET-2018-Ph-892: NEET 2018 Physics - Class 11 NEET-2018-Ph-892 Solution: A sample of 0.1 g of water at \(100^ C\) and normal pressure…

A sample of 0.1 g of water at

100^\circ C

and normal pressure (

1.013 \times 10^5\ \text{N m}^{-2}

) requires 54 cal of heat energy to convert to steam at

100^\circ C

. If the volume of the steam produced is 167.1 cc the change in internal energy of the sample is 104.3 J 208.7 J 42.2 J 84.5 J

Step-by-Step Verified Solution

Using first law of thermodynamics, _ Q = _ U + _ W _ 54 _ 4.18 = _ U + 1.013 _ 10 5 (167.1 _ 10 �6 � 0) _ _ U = 208.7 J