JEE Main · 2007 · 29 Apr 2007 · Offline · Heat & Thermodynamics

A Carnot engine, having an efficiency of

η = \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

(A)99 J
(B)90 J✓ Correct
(C)1 J
(D)100 J

Answer: (B)

Solution

A Carnot efficiency of

\frac{1}{10}

means

1 - \frac{T _{2}}{T _{1}} = \frac{1}{10}

, so

\frac{T _{2}}{T _{1}} = \frac{9}{10}

. Run backwards as a refrigerator, the work needed is

W = Q_{2} (\frac{T _{1}}{T _{2}} - 1)

, with

Q_{2}

the heat drawn from the cold reservoir. Since

\frac{T _{1}}{T _{2}} = \frac{10}{9}

, this is

W = \frac{Q _{2}}{9}

, and setting

W = 10

J gives

Q_{2} = 90

J absorbed from the low-temperature reservoir.

JEE Main · 2007 · 29 Apr 2007 · Offline · Single correct option

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