Answer to Question #206450 in Molecular Physics | Thermodynamics for Yomna

Question

A refrigerator working on reversed Carnot cycle requires 0.5 KW 

per KW of cooling to maintain a temperature of -15°C. Determine the 

following:

a) COP of the refrigerator

b) Temperature at which heat is rejected and

c) Amount of heat rejected to the surroundings per KW of cooling.

Answer 1

Answer:-

$W=0.5kw\\ Q_1=1.0kw\\ T_2=-15^o C + 273 =258K\\$

a) COP = $\frac{1.0}{0.5}=2$

b) so, cop = $\frac{T_2}{T_1-T_2}$

$2=\frac{258}{T_1-258}\\ T_1=387K=114^oC$

c)Amount of heat rejected to srrounding

$Q_1=W_1+Q_2=0.5+1 \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1.5 KW$