NEET Solved Paper 2017 Question 11
Question: A Carnot engine having an efficiency of $ \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) 1 J
C) 90 J
D) 99 J
Show Answer
Answer:
Correct Answer: C
Solution:
- $ \beta =\frac{1-\eta }{\eta } $ $ =\frac{1-\frac{1}{10}}{\frac{1}{10}}=\frac{\frac{9}{10}}{\frac{1}{10}} $ $ \beta =9 $ $ \beta =\frac{Q _2}{W} $ $ Q _2=9\times 10=90,J $
BETA
