A Carnot engine has efficiency of 50%. If the temperature of sink is reduced by 40°C, its efficiency increases by 30%. The temperature of the source will be:

Option 1 - 166.7 K

Option 2 - 255.1 K

Option 3 - 266.7 K

Option 4 - 367.7 K

6 Views|Posted 6 months ago

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Answered by

Vishal Baghel | Contributor-Level 10

6 months ago

Correct Option - 3

Detailed Solution:

Efficiency $η = 50 %$

$\Rightarrow η = (1 - \frac{T_{2}}{T_{1}}) * 100 = 50$

$\Rightarrow 1 - \frac{T_{2}}{T_{1}} = \frac{50}{100} \Rightarrow 1 - \frac{T_{2}}{T_{1}} = \frac{1}{2}$ -(i)

After the change :- $1 - \frac{T_{2}^{1}}{T_{1}} = \frac{50 * (1.3)}{100} = \frac{65}{100} = . 65$ $\frac{_{}^{}}{_{}} \frac{}{} \frac{}{}$ -(ii)

$\Rightarrow 1 - (\frac{T_{2} - 40}{T_{1}}) = . 65$ --(ii)

$\Rightarrow \frac{40}{T_{1}} = . 65 - . 50 = . 15 \Rightarrow T_{1} = \frac{4000}{15} = 266.7 k$

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