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Thermodynamics — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysicsThermodynamics5 Marks
Question
A carnot engine works between ice point and steam point. It is desired to increase the efficiency by 20% by (a) making temperature of the source as constant (b) making temperature of the sink as constant. Calculate the change in temperature in the two cases. Which one of these will you prefer and why?
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Answer

$\eta=1-\frac{\text{T}_1}{\text{T}_1}=1-\frac{273}{373}=\frac{100}{373}$

Adding 20% i.e. $\big(\frac15\big)$ to the existing $\eta,$

$\eta=\frac{100}{3+3}+20\%=\frac{100}{373}+\frac15=\frac{873}{1865}$

  1. If the temperature of the sink corresponding to $\eta'$ be $\text{T}'_2$ (source at 373k), then,

Thus, $\eta'=1-\frac{\text{T}_2'}{373}$

$\frac{\text{T}_2'}{373}=1-\eta'=1-\frac{873}{1865}=\frac{992}{1865}$

$\text{T}'_2=198.4\text{K}$

$\text{T}_2'-\text{T}_2=198.4\text{K}-273\text{K}=-74.6\text{K}$

  1. If the temperature of the source corresponding to $\eta'$ be $\text{T}'_2$ (sinkb at 273K), then

$\eta'=1-\frac{\text{T}'_2}{\text{T}'_1}$ or $\frac{873}{1865}=1-\frac{273}{\text{T}'_1}$

$\frac{273}{\text{T}'_1}=1-\frac{873}{1865}=\frac{992}{1865}$

$\text{T}'_2=513\text{K}$

$\text{T}'_1-\text{T}_1=513\text{K}-373\text{K}$

$=149.25\text{K}=140^\circ\text{C}$

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