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Heat and Thermodynamics question

2017 · Q160
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Heat and Thermodynamics question

2017 · Q160

NEETPhysicsHeat and ThermodynamicsMCQ+4 / −1
A carnot engine having an efficiency of 110{1 \over {10}}101​ 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
  1. A
    90 J
  2. B
    99 J
  3. C
    100 J
  4. D
    1 J
View written solutionFree

Correct answer: A

Given, efficiency of engine, η=110\eta = {1 \over {10}}η=101​

work done on system W = 10J

Coefficient of performance of refrigerator

β=Q2W=1−ηη=1−110110=910110=9\beta = {{{Q_2}} \over W} = {{1 - \eta } \over \eta } = {{1 - {1 \over {10}}} \over {{1 \over {10}}}} = {{{9 \over {10}}} \over {{1 \over {10}}}} = 9β=WQ2​​=η1−η​=101​1−101​​=101​109​​=9

Energy absorbed from reservoir

Q2=βw{Q_2} = \beta wQ2​=βw

Q2=9×10=90 J{Q_2} = 9 \times 10 = 90\,JQ2​=9×10=90J

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