Question #161558

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?
1
Expert's answer
2021-02-05T17:06:31-0500

The work input to a refrigerator is given by the relation

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

For a Carnot refrigerator |QH| and |QC| are related to T1 and T2 by the formula

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

The coefficient of performance of a Carnot refrigerator is

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

TH = 320 K

TC = 270 K

|QC| = 415 J

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

QH=QCTHTC=415×320270=492  J|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

=16560  cycles/s= \frac{165}{60} \;cycles/s

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

W=QHQC=492415=77  JW = |Q_H| -|Q_C| \\ = 492 -415 \\ = 77 \;J

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

W×n  J/sW \times n \;J/s

Hence, the total input power required

=77×16560  J/s=212  W= \frac{77 \times 165}{60} \;J/s \\ = 212 \;W

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

k=270320270=27050=5.4k = \frac{270}{320-270} \\ = \frac{270}{50} \\ = 5.4


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Comments

Michael
11.12.23, 17:59

Perfect! Thank you!

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