Question

igerator a coefficient of performance of 2.7? An ideal reversible refrigerator keeps its inside compartment at 6°C. What is the high temperature, Th, needed to give this refrigerator a coefficient of performance of 2.7? a. 109°C b. 8°C c. 759°C d. 22°C

Answer 1

Answer:

a. 109°C

Explanation:

The COP (coefficient of performance) of heating pump and refrigerator is the ratio of the output energy and input energy. These coefficients in terms of temperature is given as follows.

The COP of the heat pump = $K = \frac{Th}{Th-Tc}$

The COP of the refrigerator = $K = \frac{Tc}{Th-Tc}$

where Th and Tc are the hot and cold absolute temperatures.

Let's first convert Tc into kelvin.

Tc = 6°C = 6+273 = 279 K

$K = \frac{Tc}{Th-Tc}$

$2.7 =\frac{279}{Th-279}$

2.7(Th -279) = 279

2.7Th-753.3 = 279

2.7Th = 1032.3

Th = 382.33 K

Th = 382.33 - 273

Th = 109°C