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

  1. 100 J
  2. 1 J
  3. 90 J
  4. 99 J

Answer (Detailed Solution Below)

Option 3 : 90 J
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Detailed Solution

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Concept:

Carnot engine works in the reversible circle known as Carnot cycle. Carnot heat engine is the engine which have maximum efficiency to convert heat energy into work.

The expression of the efficiency of the Carnot cycle engine is:

\(η=\frac{W}{Q}\)

where \(η\) efficiency of the Carnot cycle, W and Q is the work done and amount of heat given to system respectively.

Calculation:

Given,

Efficiency η = 1/10

\(β = \frac{1-η}{η}\)

\(= \frac{{1 - \frac{1}{{10}}}}{{\frac{1}{{10}}}} = \frac{{\frac{9}{{10}}}}{{\frac{1}{{10}}}}\)

β = 9

\(\rm\beta = \frac{Q_2}{W}\)

Q2 = 9 × 10 = 90 J

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