A reversible heat engine takes heat from a source at temperature T1 then
rejects it to heat sink at temperature T3. The engine drives a reversible
refrigerator which absorbs heat from a tank at temperature T2 and rejects
heat to the same sink reservoir at temperature T3. If T1 = 1000 K and T2
= 330 K, determine: (i) The temperature T3 such that heat supplied to
engine Q1 is equal to the heat absorbed by refrigerator Q2. (ii) The
efficiency of the heat engine and C.O.P. of the refrigerator.
Given T1 = 1000 k
T2 = 330 k
T3 = Q1
Find T3 = ?
E = efficiency of engine
C.O.P of refrigerator
Q2 = ?
We know that
(i) T22 = T1 × T3
3302 = 1000 × T3
T3 = 108.9 k = Q1
(ii). Q2 = Q1 × 330/1000
= 108.9 × 330/1000
= 35.937 joule
(iii). efficiency of engine
E = 1- T2/T1
= 1- 330/1000
= 0.67
C.O.P of refrigerator
1/E = 1/0.67
= 1.4925
Comments