A) 250 cal
B) 500 cal
C) 1000 cal
D) 1500 cal
Correct Answer: B
Solution :
1 g of water \[\Rightarrow \]1 cc of water Volume of liquid\[={{V}_{L}}=1\,cc={{10}^{-6}}{{m}^{3}}\] Volume of vapours \[={{V}_{V}}=1671\,cc=1671\times {{10}^{-6}}{{m}^{3}}\] \[\Rightarrow \]\[\Delta V={{V}_{V}}-{{V}_{L}}=1670\times {{10}^{-6}}{{m}^{3}}\] \[\Rightarrow \]\[W=p\Delta V={{10}^{5}}(1670\times {{10}^{-6}})=167\,J\] \[\Rightarrow \]\[W=\frac{167}{4.18}cal=40\,cal\] Further, \[Q=mL\] \[\Rightarrow \] \[Q=(1g)\,(540\,cal\,{{g}^{-1}})=540\,cal\] According to first law of thermodynamics \[Q=\Delta U+W\] \[\Rightarrow \] \[540=\Delta U+40\] \[\Rightarrow \] \[\Delta U=500\,cal\]
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