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11.Thermodynamics
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The efficiency of Carnot engine is $50\%$ and temperature of sink is $500\,K.$ If the temperature of source is kept constant and its efficiency is to be raised to $60\%$ then the required temperature of sink will be ........... $\mathrm{K}$
A
$600$
B
$500$
C
$100$
D
$400$
Solution
Given, temperature of $\operatorname{sink}, T_2=500 K$
$\text { efficiency, } \eta_1=50 \%=\frac{1}{2} \text {. }$
efficiency of a cashot engine is given by:
$\eta=1-\frac{T_2}{T_1}$
$\frac{1}{2}=1-\frac{T_2}{T_1}$
$T_1=T_2 / 2 .$
Now, temp of source $=T_1$ (s ame)
efficiency, $\eta_2=60 \%=0.6$
temp of sink required, $T_2{ }^{\prime}=$ ?
$0.6 =1-\frac{T_2{ }^{\prime}}{T_1}$
$0.6 =1-\frac{T_2{ }^{\prime}}{T_2 / 2}=1-\frac{T_2^{\prime}}{250}$
$T_2^{\prime} =0.4 \times 250$
$T_2^{\prime} =100\,K$
Standard 11
Physics
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