Question

A heat engine placed on Neptune (planet) operating between temperature \[{T_1}\] and \[{T_2}\], where \[{T_1} > {T_2}\] has efficiency \[\dfrac{1}{5}\] , when \[{T_2}\] is lowered by 60K its efficiency increase to \[\dfrac{1}{2}\]. Then \[{T_1}\] and\[{T_2}\] are respectively.

Answer 1

Hint: Learn the definition of efficiency of a Carnot engine and the formula for efficiency of the Carnot cycle. The efficiency of a Carnot engine is the fraction of net work output by the engine. Use the formula for the efficiency in terms of the temperature of the sink and the source.

Formula used:
The efficiency of a heat engine is in terms the temperature of the reservoir and sink is given by, \[\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}\] where, \[{T_2}\] is the temperature of the sink and\[{T_1}\] is the temperature of the heat reservoir.

Complete step by step answer:
We have given here a heat engine operating on the planet Neptune. Now, the thermo dynamical laws at any place in the universe are equally valid. Hence, we can use the thermodynamics’ laws in the planet Neptune also.Now, we have given the temperature of the sink is \[{T_2}\]and the temperature of the source is \[{T_1}\]. Now, the efficiency of a Carnot cycle in terms of the source and sink temperatures is given by,
\[\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}\]

Now, we have given that the efficiency of the engine is\[\dfrac{1}{5}\] hence by the condition given is,
\[\dfrac{1}{5} = 1 - \dfrac{{{T_2}}}{{{T_1}}}\]
\[\Rightarrow \dfrac{{{T_2}}}{{{T_1}}} = 1 - \dfrac{1}{5}
\Rightarrow \dfrac{{{T_2}}}{{{T_1}}} = \dfrac{4}{5}\] ….(i)
Now, if \[{T_2}\] is lowered by 60K we will have,
\[\dfrac{1}{2} = 1 - \dfrac{{{T_2} - 60}}{{{T_1}}}\]
\[\Rightarrow \dfrac{{{T_2} - 60}}{{{T_1}}} = 1 - \dfrac{1}{2} \\
\Rightarrow \dfrac{{{T_2} - 60}}{{{T_1}}}= \dfrac{1}{2}\] ……….(ii)

Dividing equation (i) and (ii) we will have,
\[\dfrac{{\dfrac{{{T_2} - 60}}{{{T_1}}}}}{{\dfrac{{{T_2}}}{{{T_1}}}}} = \dfrac{{\dfrac{1}{2}}}{{\dfrac{4}{5}}}\]
\[\Rightarrow \dfrac{{{T_2} - 60}}{{{T_2}}} = \dfrac{1}{2} \times \dfrac{5}{4} \\
\Rightarrow \dfrac{{{T_2} - 60}}{{{T_2}}}= \dfrac{5}{8}\]
\[\Rightarrow 8{T_2} - 5{T_2} = 480\]
\[\Rightarrow 3{T_2} = 480\]
\[\Rightarrow {T_2} = 160\]
Putting the value of \[{T_2} = 160\] in equation (i) ,
\[\dfrac{{160}}{{{T_1}}} = \dfrac{4}{5}\]
\[\therefore {T_1} = 40 \times 5 = 200\]
Hence, the value of \[{T_1}\] is \[200\,K\] and the value of \[{T_2}\] is \[160\,K\].

Hence, option B is the correct answer.

Note: The temperatures taken always have to be in absolute scale else the result will be incorrect. The formula for efficiency contains only \[\dfrac{{{T_2}}}{{{T_1}}}\] it is a common mistake to write the opposite of it. If you even forget the formula, always remember that the efficiency is always less than one for real engines and the temperature of sink is always lower than the source, in that way one can easily recall the formula.