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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 KC)300 KD)450 K

Answer»

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}


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