Question

Q...The efficiency of Carnot engine is . If the sink
temperature is reduced by 100°C then engine
efficiency become 2/3. Find sink temperature and
source temperature. ​

Answer 1

$\huge\mathbb{\orange{QUESTION-}}$

The efficiency of Carnot engine is . If the sink

temperature is reduced by 100°C then engine

efficiency become 2/3. Find sink temperature and source temperature.

$\huge\mathbb{\orange{SOLUTION-}}$

Let ,

Source temperature = ${\sf T_1}$

Sink temperature = ${\sf T_2}$

$\large{\bf We\:know\:that-}$

✪$\large{\boxed{\bf \pink{ \eta = \dfrac{T_1-T_2}{T_1}} }}$

$\large\underline{\underline{\bf According\:to\: Question:}}$

$\implies{\sf \eta = \dfrac{T_1-T_2}{T_1}\:\:\implies \dfrac{1}{2}}$

$\implies{\sf 1-\dfrac{T_2}{T_1}\:\:\implies \dfrac{1}{2} }$

$\implies{\bf \dfrac{T_2}{T_1}=\dfrac{1}{2} }$

$\implies{\sf \eta ' = \dfrac{T_1-T_2'}{T_1}\:\:\implies \dfrac{2}{3}}$

$\implies{\sf 1-\dfrac{T_2'}{T_1}=\dfrac{2}{3} }$

$\implies{\bf \dfrac{T_2'}{T_1}=\dfrac{1}{3} }$

✪ Temperature is reduced by 100°C

$\implies{\sf \dfrac{T_2-100}{T_1}=\dfrac{1}{3} }$

$\implies{\sf \dfrac{T_2}{T_1}-\dfrac{100}{T_1}=\dfrac{1}{3}}$

$\implies{\sf \dfrac{1}{2}-\dfrac{100}{T_1}=\dfrac{1}{3} }$

$\implies{\sf \dfrac{1}{2}-\dfrac{1}{3}=\dfrac{100}{T_1} }$

$\implies{\sf \dfrac{1}{6}=\dfrac{100}{T_1}}$

$\implies{\bf \red{T_1=600K} }$

$\huge\mathbb{\orange{ANSWER-}}$

Source temperature is ${\bf \red{600\:K}}$