Answer to Question #206445 in Molecular Physics | Thermodynamics for Yomna

Question

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.

Answer 1

Given T₁ = 1000 k

T₂ = 330 k

T₃ = Q₁

Find T_{3 =} ?

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