A Carnot engine (E) is working between two temperatures 473 K and 273 K. In a new system two engines - engine E1 works between 473 K to 373 K and engine E2 works between 373 K to 273 K. If 12, 1 and 2 are the efficiencies of the engines E, E1 and E2, respectively, then
- 12 = 1 2
- 12 = 1 + 2
- 12 1 + 2
- 12 < 1 + 2
Answer
(4) 12 < 1 + 2
Solution
Carnot efficiency: = 1 - T2T1
12 = 1 - 273473 = 200473 0.423
1 = 1 - 373473 = 100473 0.211
2 = 1 - 273373 = 100373 0.268
1 + 2 0.479 > 12
(In general 1 - 12 = (1 - 1)(1 - 2) 12 = 1 + 2 - 1 2)
12 < 1 + 2
Answer: (4)