A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36° C, calculate the coefficient of performance.
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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![**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.
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**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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1fd69c12-64ac-4e27-aeea-6962efdd7be9%2F13d2d880-49a3-4d5a-9ee8-dfdad6d9969b%2Fv6895rb_processed.png&w=3840&q=75)
Transcribed Image Text:**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.
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