An ideal gas heat engine operates in a Carnot's cycle between 227^{0} C and 127^{0} C. It absorbs 6 × 10^{4} J at high temperature. The amount of heat converted into work is

  • Option 1)

    4.8 x 10^{4} J

  • Option 2)

    2.4 x 10^{4} J

  • Option 3)

    1.6 x 10^{4} J

  • Option 4)

    1.2 x 10^{4} J

 

Answers (1)

As we learnt in 

Efficiency of a carnot cycle -

\eta =\frac{W}{Q_{1}-Q_{2}}=1-\frac{T_{2}}{T_{1}}

T_{1}\, and\, T_{2}  are in kelvin
 

- wherein

T_{1}= Source temperature

T_{2}= Sink Temperature

\left ( T_{1} > T_{2}\right )

 

 \eta (efficiency)= 1-\frac{T_{2}}{T_{1}}=\frac{W}{Q_{1}}

\Rightarrow W= \left ( 1-\frac {T_{2}}{T_{1}} \right ).Q_{1}

=(1- \frac{400}{500})\times 6 \times 10^{4}J

= \frac{1}{5} \times 6 \times 10^{4}J= 1.2 \times 10^{4}J

Correct option is 4.

 

 


Option 1)

4.8 x 10^{4} J

Incorrect

Option 2)

2.4 x 10^{4} J

Incorrect

Option 3)

1.6 x 10^{4} J

Incorrect

Option 4)

1.2 x 10^{4} J

Correct

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