A carnot engine whose sink is at 300k has an efficiency of 49%.By how much should the temperature or
Answers
The increase in temperature is 250 K.
Explanation:
Correct statement:
A carnot engine whose sink is at 300 K 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 ?
Solution:
Temperature of sink T(L) = 300 K
Original efficiency "η" = 40η =40% = 0.4
Let the Initial temperature of the source be T(H)
Using "c" = 1 - TL /TH η
0.4 = 1 - 300 / TH
TH =500 K
Now the efficiency of the engine is increased by 50% of original efficiency,
New efficiency "η" = 40 % + 20 % = 60 %
η' = 1 - TL / T'H
0.6 = 1 - 300 / T'H
OR 300 / T'H = 0.4
T 'H = 750 K
Increase in source temperature ΔT H = T'H - TH
= 750 - 500
= 250 K
Hence the increase in temperature is 250 K.
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The efficiency of carnot engine is 50% and temperature of sink is 500 k. If temperature of source is kept constant and efficiency raised to 60%, then the required temperature of the sink will be?
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