A reversible engine converts one sixth of heat absorbed at the source into work. When the temperature of the sink is reduced by $82^\circ C$, the efficiency is doubled. Find the temperature of the source and the sink.

Hint We are provided with the efficiency of the engine and a condition that if the temperature of the sink is reduced by $82^\circ C$, then the efficiency is doubled. With this information form an equation and solve it. Use the formula of efficiency of the machine to form the equation.

Complete step by step answer
The heat Efficiency of a machine is the value of ratio of output energy to input energy. Output could be any measurable energy or temperature change while input could be heat energy or fuel which can be converted to heat energy.
${\text{efficiency, }}\eta = \dfrac{{output}}{{input}}$
${\text{Percentage efficiency, }}\eta = \dfrac{{output}}{{input}} \times 100$
For example, consider a heat machine with initial temperature T1 and final temperature as T2, then if initial temperature is greater than the final temperature and when some amount of heat is being converted into work, then T2 will be low
The efficiency of the machine is given by,
${\text{efficiency, }}\eta = \dfrac{{{T_1} - {T_2}}}{{{T_1}}}$
${\text{Percentage efficiency, }}\eta = \dfrac{{{T_1} - {T_2}}}{{{T_1}}} \times 100$
According to II law of thermodynamics, it is impossible to have a machine that converts the input energy completely into output energy or output work without any amount of heat or energy being absorbed by the machine.
The efficiency cannot be $100\% $
Given,
The initial temperature i.e. the temperature of the source is ${T_1}$
The final temperature i.e. the temperature of the sink is ${T_2}$
The given statement is
A reversible engine converts one sixth of heat absorbed at the source into work.
This implies the efficiency of the engine is $\dfrac{1}{6}$
$ \Rightarrow \eta = \dfrac{1}{6}$
We know that,
$ \Rightarrow \eta = \dfrac{{{T_1} - {T_2}}}{{{T_1}}}$
So, $ \Rightarrow \dfrac{{{T_1} - {T_2}}}{{{T_1}}} = \dfrac{1}{6}$
\[ \Rightarrow 6\left( {{T_1} - {T_2}} \right) = {T_2}\]
\[ \Rightarrow 6{T_1} - 6{T_2} = {T_2}\]
\[ \Rightarrow 6{T_1} = {T_2} + 6{T_2}\]
\[ \Rightarrow 6{T_1} = 7{T_2}\]
\[ \Rightarrow {T_1} = \dfrac{7}{6}{T_2}\]
\[ \Rightarrow {T_1} = 1.17({T_2}{\text{) }} \to {\text{1}}\]
When the temperature of the sink is reduced by $82^\circ C$, the efficiency is doubled.
$ \Rightarrow 2\eta = \dfrac{{{T_1} - ({T_2} - 82)}}{{{T_1}}}$
We know that $\eta = \dfrac{1}{6}$
$ \Rightarrow 2 \times \dfrac{1}{6} = \dfrac{{{T_1} - ({T_2} - 82)}}{{{T_1}}}$
\[ \Rightarrow \dfrac{{{T_1}}}{3} = {T_1} - ({T_2} - 82)\]
\[ \Rightarrow {T_1} = 3\left( {{T_1} - ({T_2} - 82)} \right)\]
\[ \Rightarrow {T_1} = \left( {3{T_1} - 3{T_2} + 246} \right)\]
\[ \Rightarrow - 2{T_1} + 3{T_2} = 246\]
From equation 1 substitute the value of ${T_1}$
\[ \Rightarrow - 2(1.17{T_2}) + 3{T_2} = 246\]
\[ \Rightarrow - 2.34{T_2} + 3{T_2} = 246\]
\[ \Rightarrow 0.66{T_2} = 246\]
\[ \Rightarrow {T_2} = \dfrac{{246}}{{0.66}}\]
\[ \Rightarrow {T_2} = 372.72^\circ C\]
\[ \Rightarrow {T_2} = 372.7C\]
The temperature of the source ${T_2}$ is \[372.7^\circ C\]
From equation 1 we can find ${T_1}$, the temperature of the sink.
\[ \Rightarrow {T_1} = 1.17({T_2}{\text{) }} \to {\text{1}}\]
\[ \Rightarrow {T_1} = 1.17(372.7{\text{)}}\]
\[ \Rightarrow {T_1} = 436.04^\circ C\]
\[ \Rightarrow {T_1} = 436^\circ C\]
The temperature of the source ${T_1}$ is \[436^\circ C\]
Hence,
The temperature of the source ${T_2}$ is \[372.7^\circ C\]
The temperature of the source ${T_1}$ is \[436^\circ C\]

Note In the question itself the temperature is given in Celsius so you can give the answer in Celsius. If you want to give the answer in SI unit i.e. kelvin, then you have to change the temperature given in the question to kelvin then you will get an answer in kelvin.