What is the maximum amount of work that a Carnot engine can perform per kilocalorie of a heat input if it absorbs heat at $427^\circ C$ and exhaust heat at $177^\circ C$ ?
Answer
568.5k+ views
Hint: In order to solve this question we need to understand the second law of thermodynamics and Carnot engine. So according to the second law of thermodynamics, change in entropy is defined as the amount of change in heat absorbed or expelled per second, whereas entropy is defined as disorderness in a system. Carnot engine is a system of working substances kept between two reservoirs, one is known as source and other is known as sink. Source temperature is high in comparison to sink so that working substance can absorb heat from source and deliver it to sink, also it does the work.
Complete step by step answer:
According to the question, the source temperature is, ${T_1} = 427^\circ C$.
In kelvin we get, ${T_1} = (273 + 427)K$
${T_1} = 700K$
And the sink temperature is, ${T_2} = 177^\circ C$
In kelvin we get, ${T_1} = (273 + 177)K$
${T_1} = 450K$
Let it absorb ${Q_1}$ amount of heat from the source and reject ${Q_2}$ amount of heat to sink.Let the work done by the engine is, $W$. So from definition $W = {Q_1} - {Q_2}$.So max amount of work that Carnot engine can perform per kilocalorie of heat absorbed is,
$\eta = \dfrac{W}{{{Q_1}}}$
Putting values we get,
$\eta = \dfrac{{{Q_1} - {Q_2}}}{{{Q_1}}}$
$\Rightarrow \eta = 1 - \dfrac{{{Q_2}}}{{{Q_1}}} \to (i)$
So from Carnot theorem, we now,
$\dfrac{{{Q_2}}}{{{Q_1}}} = \dfrac{{{T_2}}}{{{T_1}}}$
Putting values in equation (i) we get,
$\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}$
Putting values we get,
$\eta = 1 - \dfrac{{450}}{{700}}$
$\Rightarrow \eta = 1 - 0.64$
$\therefore \eta = 0.36$
So the maximum amount of work that a Carnot engine can perform per kilocalorie of heat absorbed is $0.36$.
Note: It should be remembered that according to Carnot theorem, no engine working between the same source and sink’s temperature can be more efficient than Carnot engine, so Carnot engine can perform only the maximum amount of work. Also here both work and heat absorbed are in units of kilocalorie, so the final answer is dimensionless and is known as efficiency.
Complete step by step answer:
According to the question, the source temperature is, ${T_1} = 427^\circ C$.
In kelvin we get, ${T_1} = (273 + 427)K$
${T_1} = 700K$
And the sink temperature is, ${T_2} = 177^\circ C$
In kelvin we get, ${T_1} = (273 + 177)K$
${T_1} = 450K$
Let it absorb ${Q_1}$ amount of heat from the source and reject ${Q_2}$ amount of heat to sink.Let the work done by the engine is, $W$. So from definition $W = {Q_1} - {Q_2}$.So max amount of work that Carnot engine can perform per kilocalorie of heat absorbed is,
$\eta = \dfrac{W}{{{Q_1}}}$
Putting values we get,
$\eta = \dfrac{{{Q_1} - {Q_2}}}{{{Q_1}}}$
$\Rightarrow \eta = 1 - \dfrac{{{Q_2}}}{{{Q_1}}} \to (i)$
So from Carnot theorem, we now,
$\dfrac{{{Q_2}}}{{{Q_1}}} = \dfrac{{{T_2}}}{{{T_1}}}$
Putting values in equation (i) we get,
$\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}$
Putting values we get,
$\eta = 1 - \dfrac{{450}}{{700}}$
$\Rightarrow \eta = 1 - 0.64$
$\therefore \eta = 0.36$
So the maximum amount of work that a Carnot engine can perform per kilocalorie of heat absorbed is $0.36$.
Note: It should be remembered that according to Carnot theorem, no engine working between the same source and sink’s temperature can be more efficient than Carnot engine, so Carnot engine can perform only the maximum amount of work. Also here both work and heat absorbed are in units of kilocalorie, so the final answer is dimensionless and is known as efficiency.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

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

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

