Answer to Question #161558 in Molecular Physics | Thermodynamics for bibo

Question

A Carnot refrigerator is operated between two heat reservoirs at temperatures of 320 K

and 270 K.

• If in each cycle the refrigerator receives 415 J of heat energy from the reservoir at 270 K, how many joules of heat energy does it deliver to the reservoir at 320 K?
• If the refrigerator completes 165 cycles each minute, what power input is required to operate it?
• What is the coefficient of performance of the refrigerator?

Answer 1

The work input to a refrigerator is given by the relation

$W = |Q_H|-|Q_C|$ (1)

For a Carnot refrigerator |Q_H| and |Q_C| are related to T₁ and T₂ by the formula

$\frac{|Q_H|}{|Q_C|} = \frac{T_H}{T_C}$ (2)

The coefficient of performance of a Carnot refrigerator is

$k = \frac{T_C}{T_H-T_C}$ (3)

T_H = 320 K

T_C = 270 K

|Q_C| = 415 J

The amount of heat delivered to the high temperature reservoir is, by using equation (2), given by

$|Q_H| = |Q_C| \frac{T_H}{T_C} \\ = \frac{415 \times 320}{270} \\ = 492 \;J$

The number of cycles per minute is n = 165 cycles/min

$= \frac{165}{60} \;cycles/s$

Using equation (1), the work input required to run the refrigerator is

$W = |Q_H| -|Q_C| \\ = 492 -415 \\ = 77 \;J$

The power input required to operate the refrigerator for n cycle per second is

$W \times n \;J/s$

Hence, the total input power required

$= \frac{77 \times 165}{60} \;J/s \\ = 212 \;W$

Using equation (3), the coefficient of performance of the Carnot refrigerator is

$k = \frac{270}{320-270} \\ = \frac{270}{50} \\ = 5.4$