Question

in a carnot s engine, when T2=0°C and T1=200°C. it's efficiency is n1. when T1=0°C and T2=-200°C. it's efficiency is n2. then what is n1/n2​

Answer 1

Part 1 :

T2 = 0°C =0+273=273K

T 1 = 200 °C=200+273=473K

$n \:1 = 1 - \frac{t2}{t1}$

$n = 1 - \frac{273}{473}$

$n 1= \frac{473 - 273}{473}$

$n 1= \frac{200}{473}$

$n1 = 0.422$

Part. 2 :

T1 = 0°C =0+273=273K

T 2= - 200 °C= -200+273= 73 K

$n \:2 = 1 - \frac{73}{273}$

$n 2= \frac{200}{273}$

$n 2= 0.73$

ratio of n 1 and n 2

$\frac{n1}{n2} = \frac{0.42}{0.73}$