A Carnot engine with efficiency 50% takes heat from a source at 600 K. In order to increase the efficiency to 70%, keeping the temperature of the sink same, the new temperature of the source will be [2023]

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Solution

Q.
A Carnot engine with efficiency 50% takes heat from a source at 600 K. In order to increase the efficiency to 70%, keeping the temperature of the sink same, the new temperature of the source will be [2023]

1 900 K

2 300 K

3 1000 K

4 360 K

Ans.
(3)

$T_{L} same$

$η = 1 - \frac{T_{L}}{T_{H}}$

$\frac{50}{100} = 1 - \frac{T_{L}}{600}$

$\frac{70}{100} = 1 - \frac{T_{L}}{T_{H}}$

$\frac{T_{L}}{600} = \frac{1}{2}, \frac{T_{L}}{T_{H}} = \frac{3}{10}$

$T_{H} = 1000 K$

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