Question

WEET
A Carnot engine has an efficiency of 20%. When
temperature of sink is reduced by 80°C, its
efficiency is doubled. The temperature of source is
[NCERT Pg. 313]​

Answer 1

Answer:

The temperature of the source is 1492°C

Explanation:

Given :

• A Carnot engine has an efficiency of 20%
• When  temperature of sink is reduced by 80°C, its  efficiency is doubled.

To find :

the temperature of source

Solution :

The formula for efficiency of a Carnot engine is given by,

    $\underline{\boxed{\bf \eta=1-\dfrac{T_2}{T_1}}}$

where

T₂ denotes the temperature of sink (in K)

T₁ denotes the temperature of source (in K)

Case - (i) :

efficiency = 20%

   $\sf 20\% =1-\dfrac{T_2}{T_1} \\\\ \sf \dfrac{20}{100}=1-\dfrac{T_2}{T_1} \\\\ \sf \dfrac{1}{5}=1-\dfrac{T_2}{T_1} \\\\ \sf \dfrac{T_2}{T_1}=1-\dfrac{1}{5} \\\\ \sf \dfrac{T_2}{T_1}=\dfrac{5-1}{5} \\\\ \sf \dfrac{T_2}{T_1}=\dfrac{4}{5} \\\\ \longrightarrow \boxed{\sf 5T_2=4T_1}$

Case - (ii)

 efficiency is doubled, = 2(20%) = 40%

temperature of sink is reduced by 80°C, 80°C = 80 + 273 K = 353 K

 T₂' = T₂ - 353

temperature of source is same i.e., T₁  

  $\sf 40\%=1-\dfrac{T_2'}{T_1} \\\\ \sf \dfrac{40}{100}=1-\dfrac{T_2 - 353}{T_1} \\\\ \sf \dfrac{2}{5}=\dfrac{T_1 -T_2 +353}{T_1} \\\\ \sf 2(T_1)=5(T_1 -T_2 +353) \\\\ \sf 2T_1=5T_1 - 5T_2 + 1765 \\\\ \sf 5T_1 - 2T_1=5T_2 - 1765 \\\\ \sf 3T_1=5T_2-1765$

  3T₁ = 4T₁ - 1765

   4T₁ - 3T₁ = 1765

     T₁ = 1765 K

     T₁ = 1765 - 273°C

     T₁ = 1492°C  

Therefore, the temperature of the source is 1492°C