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A Carnot engine, whose efficiency is 40%, takes in heat from a source maintained at a temperature of 500 K It is desired to have an engine of efficiency 60%. Then, the intake temperature for the same exhaust (sink) temperature must be

Option: 1

efficiency of Carnot engine cannot be made larger than 50%


Option: 2

1200 K


Option: 3

750 K


Option: 4

600 K


Answers (1)

best_answer

As learnt 

Efficiency of Carnot engine,

\eta =1-\frac{T_{2}}{T_{1}}     where  T_{1}  is the temperature of the source and T_{2} is the temperature of the sink.

Case I:

\\ \\ \eta =0.4 \\ \\ T_{1}=500\: K \\ \\ T_{2}=constant

n=1 -\frac{T_{2}}{T_{1}} \: \: or \: \: 0.4=1-\frac{T_{2}}{500}

or \: \: \frac{T_{2}}{500}=0.6 \: \: \: \: \: \Rightarrow T_{2}=300K

 

Case II: If efficiency becomes 0.6

\Rightarrow 0.6 \: \: \: =1-\frac{T_{2}}{T_{1}} \: \: \: = 1-\frac{300}{T_{1}}

or \: \frac{300}{T_{1}}=0.4

or \: \: \: T_{1}=\frac{300}{0.4}=750K

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manish

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