A reversible heat engine converts one-fourth of the heat input into work. When the temperature of the sink is reduced by 52 K, its efficiency is doubled. The temperature in Kelvin of the source will be ______. 

CONCEPT:

• The efficiency of the reversible heat engine is written as;

            η = \(\frac{W}{Q_{in}}\)     ----(1)

           Here we have W is the work done, and Q_in is the heat.

• The efficiency of the reversible heat engine by a given temperature is written as;

           η = 1 - \(\frac{T_2}{T_1}\)     ----(2)

          Here we have T₁ as the temperature of the source and T₂ as the temperature of the sink. Here T₁ < T₂

CALCULATIONS:

Given: Efficiency, η = \(\frac{W}{Q_{in}}\) = \(\frac{1}{4}\)      ----(3)

The efficiency of the heat engine is written as;

 η = 1 - \(\frac{T_2}{T_1}\)  

Now, on putting the equation (3) above we have;

\(\frac{1}{4}\) = 1 - \(\frac{T_2}{T_1}\)  

⇒ \(\frac{T_2}{T_1}\) = 1 - \(\frac{1}{4}\)

⇒ \(\frac{T_2}{T_1}\) = \(\frac{4-1}{4}\)

⇒  \(\frac{T_2}{T_1}\) = \(\frac{3}{4}\)

Now, the temperature of the sink is reduced by 52 K, and efficiency becomes doubled, equation (2) is written as;

Now, putting all the given values we have ;

 \(\frac{1}{2}\) = 1 - \(\frac{T_2-52}{T_1}\)  

⇒ \(\frac{T_2-52}{T_1}\) = \(\frac{1}{2}\)

⇒ \(\frac{T_2}{T_1}\) - \(\frac{52}{T_1}\) = \(\frac{1}{2}\)

⇒ \(\frac{3}{4}\) - \(\frac{52}{T_1}\) = \(\frac{1}{2}\)

⇒ \(\frac{52}{T_1}\) = \(\frac{3}{4}\) - \(\frac{1}{2}\)

⇒ \(\frac{52}{T_1}\) = \(\frac{1}{4}\)

⇒ T₁ = 208 K .

Hence the temperature of the source is 208 K.