Question

The ratio $(W/Q)$ for a carnot–engine is $\frac 16$, Now the temperature of sink is reduced by $62 \ ºC$, then this ratio becomes twice, therefore the initial temperature of the sink and source are respectively:

• A. $33°C,\ 67°C$
• B. $37°C,\ 99°C$
• C. $67°C ,\ 33°C$
• D. $97K,\ 37K$

Answer 1

Answer: (B)

$37°C,\ 99°C$

Answer 2

$\frac WQ=\frac 16$

$1-\frac {T_L}{T_H}=\frac 16$

$\frac {T_L}{T_H}=\frac 56$

$If\ sink \ temp. \ decrease\ by \ 62°C\ then\$

$1-\frac {T_L-62}{T_H}=\frac 26$

⇒ $\frac {T_L-62}{T_H}=\frac 23$

$2T_H=3T_L-186$

⇒ $2T_H=3×\frac 56T_H-186$

$2T_H-\frac 52T_H=-186$

⇒ $\frac {5-4}{2}T_H=186$

$T_H=186×2 =372K=99°C$

$TL=\frac 56×372=310K=37°C$

So, \ the\ correct \ option\ is\ (B): $$37°C,\ 99°C