How much work must be done to extract 15 J of heat from a reservior at 4ºC and transfer it to one at 24ºC by means of a refrigerator using a Carnot cycle?

CONCEPT:

Refrigerator:

• A refrigerator is a device that maintains a system or a body at a temperature lower than the surrounding temperature.
• In the refrigerator, the heat is removed from the low-temperature body and is rejected to the high-temperature body.
• Generally, the surrounding atmosphere acts as the high-temperature body for a refrigerator.

​Coefficient of Performance:

• The coefficient of performance is defined as the ratio of the desired output to the work input.
• The coefficient of performance of a refrigerator is given as,

\(⇒ COP_{R}=\frac{Q_{L}}{W}=\frac{Q_{L}}{Q_{H}-Q_{L}}=\frac{T_{L}}{T_{H}-T_{L}}\)

Where QL = heat removed from the system, QH = heat given to the surrounding, W = work input, TL = temperature of the system, and TH = temperature of the surrounding

CALCULATION:

Given Q_L = 15 J, T_L = 4°C = 4 + 273 K, and TH = 24°C = 24 + 273 K

• We know that the coefficient of performance of a refrigerator is given as,

\(⇒ COP_{R}=\frac{Q_{L}}{W}=\frac{T_{L}}{T_{H}-T_{L}}\)

\(⇒ \frac{15}{W}=\frac{4+273}{(24+273)-(4+273)}\)

⇒ W = 1.08 J

• Hence, option 3 is correct.