Question #303799

A Carnot engine absorbs heat from a source at a higher temperature of 800 Kelvin

and the efficiency of the Carnot engine is 50%.

To what value the higher temperature should be increased for efficiency to increase

to 80%, if the lower temperature is kept constant.


1
Expert's answer
2022-03-07T14:56:03-0500


η\eta =t2t1t2%=\frac{t_2-t_1}{t_2}\% or (1T2T1)(1-\frac{T_2}{T_1})


Where t2t_2 is the source temperature

t1=t_1= intake temperature


50100=800t1800    t1=400K\frac{50}{100}=\frac{800-t_1}{800}\implies\>t_1=400K


Where the efficiency is 80%80\%


80100=t2400t2\frac{80}{100}=\frac{t_2-400}{t_2}


t2=2000Kt_2=2000K





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