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Thermodynamics — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsThermodynamics5 Marks
Question
A Carnot engine is working between ice point and steam point. It is desired to increase its efficiency by 20% (a) by changing temperature of hot reservoir alone, (b) by changing temperature of colder reservoir only. Calculate the change in temperature in each case.
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Answer

Here, T1 = 100°C = 373K and T2 = 0°C = 273K

$\eta=\frac{\text{T}_1-\text{T}_2}{\text{T}_1}=\frac{373-273}{373}$

$=\frac{100}{373}=0.268$

As we want to increase its efficiency by 20%, hence new efficiency is,

$\eta'=26.\%+20\%=46.8\%$

  1. If keeping temperature of colder reservoir fixed the temperature of hot reservoir is changed to $\text{T}_1'$ then

$46.8=\frac{\text{T}_1'-273}{\text{T}'_1}\times100$

$\Rightarrow46.8\text{T}_1'=100$

$\text{T}_1'-27300$

$\Rightarrow53.2\text{T}'_2=27300$ or $\frac{27300}{53.2}=\text{513.2}\text{K}$

$\therefore\text{T}_1'-\text{T}_1=513.2-373=140.2\text{K}$

It means that temprature of hot resevoir be raised by 140.2K

  1. If keeping temperature of hot reservoir fixed, the temperature of colder reservoir is changed to $\text{T}_2',$ then

$46.8=\frac{\text{T}_1-\text{T}_2'}{\text{T}_1}\times100$

$=\frac{373-\text{T}'_2}{373}\times100$

$\therefore373\times46.8=373\times100-100\text{T}_2'$

$\Rightarrow100\text{T}_2'=373\times(100-46.8)$

$=373\times53.2$

$\Rightarrow\text{T}'_2=\frac{373-53.2}{100}=198.4\text{K}$

$\therefore\text{T}_2-\text{T}_2'=273-198.4$

$=74.6\text{K}=74.6^\circ\text{C.}$

It means that temparture of colder reservoir be lowered by 74.6°C.

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