**Problem Statement:**A refrigerator is to maintain eatables kept inside at 9°C. If the room temperature is 36°C, calculate the coefficient of performance.---**Solution:**To solve this problem, we need to determine the Coefficient of Performance (COP) for the refrigerator. The COP of a refrigerator is given by the formula:\[ \text{COP} = \frac{T_{\text{low}}}{T_{\text{high}} - T_{\text{low}}} \]where:- \( T_{\text{low}} \) is the temperature inside the refrigerator in Kelvin.- \( T_{\text{high}} \) is the temperature outside (room temperature) in Kelvin.First, we convert the given temperatures from Celsius to Kelvin by adding 273 to each temperature in Celsius.\[ T_{\text{low}} = 9°C + 273 = 282 \, \text{K} \]\[ T_{\text{high}} = 36°C + 273 = 309 \, \text{K} \]Now substitute the values into the COP formula:\[ \text{COP} = \frac{282 \, \text{K}}{309 \, \text{K} - 282 \, \text{K}} = \frac{282}{27} \approx 10.44 \]Therefore, the coefficient of performance (COP) of the refrigerator is approximately 10.44.

Given, Temperature required inside refrigerator = TL = 9 oC = (273+9) K = 282 K Temperature of…

Answered: A refrigerator is to maintain eatables…

A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36° C, calculate the coefficient of performance.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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