A Carnot engine having an efficiency of 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 :
Answer & explanation
Correct answer: option 3
Work done = Q2(\(\frac{T_1}{T_2}\)-1) -------(i)
η = 1 - \(\frac{T_2}{T_1}\)
\(\frac{1}{10}\) = 1 - \(\frac{T_2}{T_1}\)
\(\frac{T_2}{T_1}\) = 1 - \(\frac{1}{10}\)
\(\frac{T_2}{T_1}\) = \(\frac{9}{10}\)
⇒ \(\frac{T_1}{T_2}\) = \(\frac{10}{9}\)
Using equation 1
10 = Q2(\(\frac{10}{9}\)-1)
Q2 = 90J