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Andrej [43]
3 years ago
10

Please help me and i will mark u as brainlist

Physics
1 answer:
forsale [732]3 years ago
8 0

Answer:

a) 8 secs I think

b)2m/s^2

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What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a
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Complete Question

A certain refrigerator, operating between temperatures of -8.00°C and +23.2°C, can be approximated as a Carnot refrigerator.

What is the refrigerator's coefficient of performance? COP

(b) What If? What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a heat pump? COP

Answer:

a

 COP = 8.49

b

  COP_1 = 9.49  

Explanation:

From the question we are told that

     The lower operation temperature of refrigerator is  T_1 =  -8.00^oC =  265 \  K

     The upper operation temperature of the refrigerator is   T_2 =  23.2 ^oC =  296.2 \  K

Generally the refrigerators coefficient of performance is mathematically represented as

        COP =  \frac{T_1}{T_2 - T_1  }

=>     COP =  \frac{265}{296.2 - 265  }

=>     COP = 8.49

Generally if a refrigerator (operating between the same temperatures) was instead used as a heat pump , the coefficient of performance is mathematically represented as

            COP_1 =  \frac{T_2}{ T_2 - T_1}  

=>         COP_1 =  \frac{296.2}{ 296.2 - 265 }  

=>         COP_1 = 9.49  

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