Neet Solved Paper 2015 Question 20
Question: A Carnot engine, having an efficiency of $ \eta =\frac{1}{10} $ as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is
Options:
A) $ 100,J $
B) $ 99J $
C) $ 90J $
D) $ 1J $
Show Answer
Answer:
Correct Answer: C
Solution:
- As, $ Q _1+W=Q _2 $ Given, $ \eta =\frac{1}{10} $ Now, using $ \eta =1-\frac{T _1}{T _2} $ So, $ \frac{1}{10}=1-\frac{T _1}{T _2} $
$ \Rightarrow ,\frac{T _1}{T _2}=\frac{9}{10} $ Now $ \frac{Q _1}{Q _2}=\frac{T _1}{T _2} $
$ \Rightarrow ,\frac{Q _1}{Q _1+W}=\frac{9}{10} $
$ \Rightarrow $ $ 10Q _1,=9Q _1,+9W $
$ \Rightarrow Q _1=9W=9\times 10=90J $
