NEET Solved Paper 2018 Question 45
Question: A sample of 0.1 g of water at $ \text{100 }{}^\circ\text{ C} $ and normal pressure $ \text{(1}\text{.013 }\times\text{ 1}{0^{5}}N{m^{\text{–2}}}\text{)} $ requires 54 cal of heat energy to convert to steam at $ \text{100 }{}^\circ\text{ C} $ . If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is [NEET - 2018]
Options:
A) 42.2 J
B) 208.7 J
C) 104.3 J
D) 84.5 J
Show Answer
Answer:
Correct Answer: B
Solution:
$ \Delta \text{Q=}\Delta \text{U+}\Delta W $
$ \Rightarrow 54\times 4.18=\Delta U+1.013\times 10^{5}(167.1\times {10^{-6}}-0) $
$ \Rightarrow \Delta U=208.7J $
BETA
