Question

Question 1 : A carnot engine whose sink is at 300K has an efficiency of 40%. By how much should the Temperature of source be increased to as to increase its efficiency by 50% of original efficiency ?
A)150 K
B) 250 K
C)300 K
D)450 K

Answer 1

Temperature of sink ($T_c$) = 300K
Temperature of source = $T_h$

Initial efficiency = 40%
$efficiency,\ \eta_1=1- \frac{T_c}{T_h_1} \\ \\0.4=1- \frac{300}{T_h_1} \\ \\ \frac{300}{T_h_1} =1-0.4=0.6\\ \\T_h_1= \frac{300}{0.6}=500K$

If we need to increase the efficiency by 50% of original,
efficiency = 40% + (50/100)×40% = 40%+20% = 60%

$efficiency,\ \eta_2=1- \frac{T_c}{T_h_2} \\ \\0.6=1- \frac{300}{T_h_2} \\ \\ \frac{300}{T_h_2} =1-0.6=0.4\\ \\T_h_2= \frac{300}{0.4}=750K$

Change in temperature = $750K-500K=\boxed{250K}$

Answer 2

Answer:

See the attachment

Good luck

Explanation: