
At what temperature must an ideal heat engine absorb heat if its efficiency is $38\% $, when it exhausts heat at a temperature of $60^\circ C$?
A. $533^\circ C$
B. $264^\circ C$
C. $264K$
D. $96.8K$
Answer
497.1k+ views
Hint: The efficiency of an engine only depends on the temperatures of the hot and cold bodies between which the engine works. Using the equations of Carnot’s engine and applying the formula which equates efficiency with the temperature of the cold body and the hot body, this question can be easily solved.
Formula used :
$\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}$ ,where$\eta $=efficiency, ${T_2}$=Lower temp (exhaust temp in case of heat engine), ${T_1}$=higher temp (absorbed heat temp in case of heat engine)
Complete step by step solution:
Let the temp at which the engine absorbs heat be ${T_A}$.
Now we know that the Carnot’s engine’s efficiency formula is
$\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}$.
Utilising the above formulae in the given question, we change the variables and rename them as ${T_2} = {T_E}$ and ${T_1} = {T_A}$.
Therefore, utilising the above equation in this question, we can transform it to -
This is the required equation to find out the temperature at which the Carnot’s engine absorbs heat.
Now
$
\Rightarrow \eta = 1 - \dfrac{{{T_E}}}{{{T_A}}} \\
\Rightarrow \eta + \dfrac{{{T_E}}}{{{T_A}}} = 1 \\
\Rightarrow \dfrac{{{T_E}}}{{{T_A}}} = 1 - \eta \\
\Rightarrow \dfrac{{{T_A}}}{{{T_E}}} = \dfrac{1}{{1 - \eta }} \\
\Rightarrow {T_A} = \dfrac{{{T_E}}}{{1 - \eta }} \\
$,
substituting the respective values in their respective places, and equating the above equation, we get -
$
\Rightarrow {T_A} = \dfrac{{{T_E}}}{{1 - \eta }} \\
\Rightarrow {T_A} = \dfrac{{333K}}{{1 - 0.38}} \\
\Rightarrow {T_A} = \dfrac{{333K}}{{0.62}} \\
\Rightarrow {T_A} = 537K \\
$
Since the answer in Kelvin does not match any of the above options, we convert it to Celsius.
Now, converting the temp from Kelvin to Celsius, we have -
$537K = (537 - 273)^\circ C = 264^\circ C$
Therefore, option (B) is the correct answer .
Note:
While using the values of temperature in the given formula, make sure to convert them to Kelvin. Using celsius values will lead to incorrect answers. In the given question, we are dealing with an ideal engine (an engine in which no energy is lost). But, in the real world, engines lose energy through friction and turbulence.
Formula used :
$\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}$ ,where$\eta $=efficiency, ${T_2}$=Lower temp (exhaust temp in case of heat engine), ${T_1}$=higher temp (absorbed heat temp in case of heat engine)
Complete step by step solution:
Let the temp at which the engine absorbs heat be ${T_A}$.
Now we know that the Carnot’s engine’s efficiency formula is
$\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}$.
Utilising the above formulae in the given question, we change the variables and rename them as ${T_2} = {T_E}$ and ${T_1} = {T_A}$.
Therefore, utilising the above equation in this question, we can transform it to -
This is the required equation to find out the temperature at which the Carnot’s engine absorbs heat.
Now
$
\Rightarrow \eta = 1 - \dfrac{{{T_E}}}{{{T_A}}} \\
\Rightarrow \eta + \dfrac{{{T_E}}}{{{T_A}}} = 1 \\
\Rightarrow \dfrac{{{T_E}}}{{{T_A}}} = 1 - \eta \\
\Rightarrow \dfrac{{{T_A}}}{{{T_E}}} = \dfrac{1}{{1 - \eta }} \\
\Rightarrow {T_A} = \dfrac{{{T_E}}}{{1 - \eta }} \\
$,
substituting the respective values in their respective places, and equating the above equation, we get -
$
\Rightarrow {T_A} = \dfrac{{{T_E}}}{{1 - \eta }} \\
\Rightarrow {T_A} = \dfrac{{333K}}{{1 - 0.38}} \\
\Rightarrow {T_A} = \dfrac{{333K}}{{0.62}} \\
\Rightarrow {T_A} = 537K \\
$
Since the answer in Kelvin does not match any of the above options, we convert it to Celsius.
Now, converting the temp from Kelvin to Celsius, we have -
$537K = (537 - 273)^\circ C = 264^\circ C$
Therefore, option (B) is the correct answer .
Note:
While using the values of temperature in the given formula, make sure to convert them to Kelvin. Using celsius values will lead to incorrect answers. In the given question, we are dealing with an ideal engine (an engine in which no energy is lost). But, in the real world, engines lose energy through friction and turbulence.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

