A cornot engine whose heat sinks at 27°C has an efficiency of 25%. By how many degrees should the temperature of the source be changed to increase the efficiency by 100% of the original efficiency?

Option 1 -

Increases by 18°C

Option 2 -

Increases by 200°C

Option 3 -

Increases by 120°C

Option 4 -

Increases by 75°C

2 Views | Posted 2 months ago

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

0.25 = 1 - $\frac{T_{2}}{T_{1}}$ $\frac{_{}}{_{}}$

0.25 = 1 - $\frac{(27 + 273)}{T_{1}} \Rightarrow \frac{300}{T_{1}} = 1 - 0.25$ $\frac{}{_{}} \frac{}{_{}}$

$\Rightarrow \frac{300}{T_{1}} = 0.75$

$T_{1} = \frac{300 \times 100}{75}$

T1 = 400 k

Now, efficiency increases by 100%

$η_{2} = 100 % η_{1} + η_{1}$

$= 2 η_{1}$

= 0.25 × 2

= 0.50

0.50 = 1 - $\frac{T_{2}'}{T_{1}}$ $\frac{_{}}{_{}}$

$\Rightarrow \frac{T_{2}'}{T_{1}} = 0.50$

$T_{1}' = \frac{300}{0.5}$

$T_{1}' = 600 k$

$T_{1}' - T_{1} =$ (600 – 400) = 200 k or 200° C

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A cornot engine whose heat sinks at 27°C has an efficiency of 25%. By how many degrees should the temperature of the source be changed to increase the efficiency by 100% of the original efficiency?

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