Question

Thermodynamics

two part question

Question 15 1 pts Problems 15 and 16 are based on the following description A household refrigerator runs 20 minutes in every hour and removes heat from the food compartment during this running time at a rate of 1200 kJ/h. The COP of the refrigerator is 2.1. The power the refrigerator draws when running in kW is closer to: O 0.16 O 0.23 O 0.66 O 0.82 O 0.08 0.27 21 O 11 Question 16 1 pts Assuming a cost of electricity of 0.10 $/kWh, the cost in $ to operate this refrigerator per day is closer to:

Answer 1

Use the expression of COP to calculate the power required to operate the refrigerator. Find out the total time in a day when refrigerator is working and change this to KWh to calculate the total cost of operation in a day.

Solution U Cooling rate (RO)=) 1200 kJ h RC = 1200 KI - 0.333 KW 360os COP » ☺ R.C Winput y power by re required repigerator Winput 2.1 > 0.333 Winput Winput » 0.1587 Ary Winput a 0.16 kW works for 20x 24 minutes ore foi gerator In day one -> 480 minutes 8 hours. Winput for one day 0.16 x 8 kWh 1.28 kWh I kWh For cost is → 0.10$ For 1.28 kWh kwh cost is 0.16 x 1.28 Cost => $0.128. Ang