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JEE Main 20208 JanuaryEvening Shift
Exam
JEE Main
Year
2020
Date
8 January
Shift
Evening Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2020 · 8 January · Evening Shift · Physics · Thermodynamics

A Carnot engine having an efficiency of (1)/(10) is being used as a...

A Carnot engine having an efficiency of 110\dfrac{1}{10} is being used as a refrigerator. If the work done on the refrigerator is 10J,10J, the amount of heat absorbed from the reservoir at a lower temperature is

Options

  1. A99J99J
  2. B100J100J
  3. C1J1J
  4. D90J90J

Correct answer

D. 90J90J

Step-by-step solution

For Carnot engine using as refrigerator
W=Q2(T1T2−1)W = {Q}_{2}\left(\dfrac{{T}_{1}}{{T}_{2}} - 1\right)
It is given η=110⇒η=1−T2T1⇒T2T1=910η = \dfrac{1}{10} ⇒ η = 1 - \dfrac{{T}_{2}}{{T}_{1}} ⇒ \dfrac{{T}_{2}}{{T}_{1}} = \dfrac{9}{10}
So ,Q2=90J(asW=10J), {Q}_{2} = 90J\left(asW = 10J\right)
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