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# A Carnot engine shows efficiency 40% on the energy at 500 K. To increase the efficiency to 50 at what temperature it should take energy.A) 400 KB) 700 KC) 600 KD) 800 K

Last updated date: 13th Jun 2024
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Hint: According to Kelvin-Planck statement for an engine to produce work the engine should come in contact with a source and also should come in contact with a sink temperature and accordingly the Carnot engine when comes in contact with source and sink temperature it produces work.

Formula Used: The formula of the efficiency of an engine is given by,
$\Rightarrow \eta = \left( {1 - \dfrac{{{T_2}}}{{{T_1}}}} \right) \times 100$
Where $\eta$ is the efficiency in percentage the temperature of source is ${T_1}$ and temperature of ${T_2}$ is the temperature of sink.

It is given in the problem that a Camot engine shows efficiency 40% on the energy at 500 K and we need to find the temperature of source if the efficiency is 50%.
Let us first of all calculate the temperature of sink i.e. ${T_2}$.
The efficiency of the engine is given as 40% and the source temperature is 500 K, therefore.
$\Rightarrow \eta = \left( {1 - \dfrac{{{T_2}}}{{{T_1}}}} \right) \times 100$
$\Rightarrow 40 = \left( {1 - \dfrac{{{T_2}}}{{500}}} \right) \times 100$
$\Rightarrow 0 \cdot 40 = \left( {1 - \dfrac{{{T_2}}}{{500}}} \right)$
$\Rightarrow 0 \cdot 40 = \left( {\dfrac{{500 - {T_2}}}{{500}}} \right)$
$\Rightarrow \left( {0 \cdot 40 \times 500} \right) = 500 - {T_2}$
$\Rightarrow {T_2} = 500 - \left( {0 \cdot 40 \times 500} \right)$
$\Rightarrow {T_2} = 500 - 200$
$\Rightarrow {T_2} = 300k$
The sink temperature is 300 k.
Now if the efficiency is 50 % let us then calculate the temperature of the source.
The formula of the efficiency of an engine is given by,
$\Rightarrow \eta = \left( {1 - \dfrac{{{T_2}}}{{{T_1}}}} \right) \times 100$
Where $\eta$ is the efficiency in percentage the temperature of source is ${T_1}$ and temperature of ${T_2}$ is the temperature of sink.
As the efficiency is 50 % and the sink temperature is 300 k.
$\Rightarrow \eta = \left( {1 - \dfrac{{{T_2}}}{{{T_1}}}} \right) \times 100$
$\Rightarrow 50 = \left( {1 - \dfrac{{300}}{{{T_1}}}} \right) \times 100$
$\Rightarrow 0 \cdot 50 = \left( {1 - \dfrac{{300}}{{{T_1}}}} \right)$
$\Rightarrow 0 \cdot 50 = \left( {\dfrac{{{T_1} - 300}}{{{T_1}}}} \right)$
$\Rightarrow 0 \cdot 50{T_1} = {T_1} - 300$
$\Rightarrow 0 \cdot 50{T_1} = 300$
$\Rightarrow {T_1} = \left( {\dfrac{{300}}{{0 \cdot 50}}} \right)$
$\Rightarrow {T_1} = 600k$
The source temperature for the efficiency to be 50% is ${T_1} = 600k$.
The correct answer for this problem is option C.

Note: The work produced can be increased by increasing the source temperature and keeping the sink temperature same or the work can also be increased if we decrease the sink temperature these are the two ways by which the efficiency can be increased and also both of the conditions can be applied along simultaneously to achieve the increase in work output and the efficiency of the engine.