JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Heat & Thermodynamics

A Carnot engine having an efficiency of

\frac{1}{10}

is being used as a refrigerator. If the work done on the refrigerator is 10 J, the amount of heat absorbed from the reservoir at lower temperature is:

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

Answer: (D)

Solution

A Carnot engine of efficiency

\frac{1}{10}

has

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

, so

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

. Run in reverse as a refrigerator between the same reservoirs, the work needed to lift heat

Q_{2}

from the cold side is

W = Q_{2} (\frac{T _{1}}{T _{2}} - 1) = Q_{2} (\frac{10}{9} - 1) = \frac{Q _{2}}{9}

. With

W = 10

Q_{2} = 9 W = 90

JEE Main · 2020 · 08 Jan 2020 · Shift 2 · Single correct option

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