A sample of `0.1 g` of water of `100^(@)C` and normal pressure `(1.013 xx 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 isA. `84.5J`B. `104.3J`C. `42.2J`D. `208.7J`

A sample of `0.1 g` of water of `100^(@)C` and normal pressure `(1.013

1.	A sample of `0.1 g` of water of `100^(@)C` and normal pressure `(1.013 xx 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 isA. `84.5J`B. `104.3J`C. `42.2J`D. `208.7J`
Answer» Correct Answer - D `Delta Q = 54cal = 54xx4.18joule = 225.72` joule `DeltaW = P[V_(steam)-V_(water)]` [for water 0.1 gram = 0.1 c c] `=1.013xx10^(5)[167xx10^(-6)-0.1xx10^(-6)]"joule"` `=1.013xx167xx10^(-1)=16.917 "joule"` By first law of thermodyamics `implies DeltaU=DeltaQ-DeltaW=225.72-16.917` `DeltaU = 208.8 "Joule"`.

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A sample of `0.1 g` of water of `100^(@)C` and normal pressure `(1.013 xx 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 isA. `84.5J`B. `104.3J`C. `42.2J`D. `208.7J`

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