Question

Imagine a Carnot engine has a hot reservoir of 680 K and a cold reservoir of 220 K. What is the efficiency of the engine? 58.3% 75.9% 62.3% 72.1% 67.6%

Answer 1

N = 1 - (QC/QH)

n = efficiency

QC = Cold Reservoir = 220 K
QH = Hot Reservoir = 680 K

$n = 1 - (220 \div 680) (\div 10)\\ n = 1 - (22 \div 68) (\div 2) \\ n = 1 - (11 \div 34) \\ n = (34 - 11) \div 34 \\ n = 23 \div 34 \\ n = 0.676... \times 100\\ n = 67.6$

Answer: The engine's efficiency is 67.6% .

Answer 2

the other person's right, it's E. 67.6% for plato users :)