Question 785: Thermodynamics & KTG

Question

A carnot engine, whose efficiency is $40 \%$ takes heat from a source maintained at temperature 600 K . It is desired to have an efficiency $60 \%$, then the intake temperature for the same exhaust (sink) temperature should be

Accepted Answer

To find the required intake temperature for a Carnot engine with increased efficiency, we can use the formula for the efficiency ($\eta$) of a Carnot engine:
$ \eta = 1 - \frac{T_{\mathrm{L}}}{T_{\mathrm{H}}} $
Initially, the given efficiency is 40%, and the source temperature ($T_{\mathrm{H}}$) is 600 K. We need to determine the sink temperature ($T_{\mathrm{L}}$):
$ \frac{40}{100} = 1 - \frac{T_{\mathrm{L}}}{600} $
Solving for $T_{\mathrm{L}}$:
$ \frac{T_{\mathrm{L}}}{600} = \frac{60}{100} $
$ T_{\mathrm{L}} = 360 \, \mathrm{K} $
Now, for the desired efficiency of 60%, we use:
$ \frac{60}{100} = 1 - \frac{360}{T_{\mathrm{H}}} $
Solving for the new intake temperature ($T_{\mathrm{H}}$), we have:
$ \frac{360}{T_{\mathrm{H}}} = \frac{40}{100} $
$ T_{\mathrm{H}} = 900 \, \mathrm{K} $
Therefore, to achieve an efficiency of 60%, the intake temperature should be 900 K.

Answer

1800 K

Answer

1200 K

Answer

900 K

Answer

600 K

Thermodynamics & KTG

A carnot engine, whose efficiency is 40 \% takes heat from a source maintained at temperature 600 K . It is desired to have an efficiency 60 \%, then the intake temperature for the same exhaust (sink) temperature should be

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