Answer
Verified
447.6k+ views
Hint: The work done by a Carnot engine is equal to the difference between the amount of heat drawn from the reservoir and the amount of heat dumped into the sink. The efficiency of a Carnot engine gives a relation between heat energy and temperature of reservoir and sink.
Formula Used:
The formula for efficiency of heat engine is given as
$\eta = 1 - \dfrac{{{Q_2}}}{{{Q_1}}} = 1 - \dfrac{{{T_2}}}{{{T_1}}}{\text{ }}...{\text{(i)}}$
where ${Q_1} = $ heat given to the Carnot engine or drawn from the reservoir
${Q_2} = $heat obtained from the Carnot engine or dumped into the sink
${T_1} = $temperature of the reservoir
${T_2} = $temperature of the sink.
Work done by a Carnot engine is given as
$W = {Q_1} - {Q_2}{\text{ }}...{\text{(ii)}}$
Complete step-by-step answer:
A Carnot engine is an ideal heat engine which converts heat energy into work. It draws heat from a reservoir at certain temperature and expels heat into a sink at certain temperature. The efficiency of a heat engine is the amount of useful work obtained from a given input energy.
We are given the following values for a Carnot engine:
Heat drawn from the reservoir: ${Q_1} = 900kcal = 9 \times {10^5}cal$
Temperature of the reservoir: ${T_1} = 723^\circ C = 723 + 273 = 996K$
Temperature of the sink: ${T_2} = 30^\circ C = 30 + 273 = 303K$
Now we first need to find the heat dumped into sink i.e. ${Q_2}$ which can done using equation (i) as follows:
$
1 - \dfrac{{{Q_2}}}{{{Q_1}}} = 1 - \dfrac{{{T_2}}}{{{T_1}}}{\text{ }} \\
\Rightarrow \dfrac{{{Q_2}}}{{{Q_1}}} = \dfrac{{{T_2}}}{{{T_1}}} \\
\Rightarrow {Q_2} = \dfrac{{{T_2}}}{{{T_1}}} \times {Q_1} \\
$
Substituting the known values, we get
$
{Q_2} = \dfrac{{303}}{{996}} \times 9 \times {10^5} \\
= 2.737 \times {10^5}cal \\
$
Now we can calculate the work done by the Carnot engine using the equation (ii) as follows:
$W = {Q_1} - {Q_2}{\text{ }}$
Substituting the values, we get
$
W = 9 \times {10^5} - 2.737 \times {10^5} \\
\simeq 6.26 \times {10^5}cal \\
$
Hence, the correct answer is option C.
Note: The Carnot cycle is reversible in nature and its efficiency is the highest possible for an engine. The ideal temperature of the sink should be 0K to get maximum efficiency but it is not possible to practically attain this temperature due to which carnot engine is considered ideal in nature.
Formula Used:
The formula for efficiency of heat engine is given as
$\eta = 1 - \dfrac{{{Q_2}}}{{{Q_1}}} = 1 - \dfrac{{{T_2}}}{{{T_1}}}{\text{ }}...{\text{(i)}}$
where ${Q_1} = $ heat given to the Carnot engine or drawn from the reservoir
${Q_2} = $heat obtained from the Carnot engine or dumped into the sink
${T_1} = $temperature of the reservoir
${T_2} = $temperature of the sink.
Work done by a Carnot engine is given as
$W = {Q_1} - {Q_2}{\text{ }}...{\text{(ii)}}$
Complete step-by-step answer:
A Carnot engine is an ideal heat engine which converts heat energy into work. It draws heat from a reservoir at certain temperature and expels heat into a sink at certain temperature. The efficiency of a heat engine is the amount of useful work obtained from a given input energy.
We are given the following values for a Carnot engine:
Heat drawn from the reservoir: ${Q_1} = 900kcal = 9 \times {10^5}cal$
Temperature of the reservoir: ${T_1} = 723^\circ C = 723 + 273 = 996K$
Temperature of the sink: ${T_2} = 30^\circ C = 30 + 273 = 303K$
Now we first need to find the heat dumped into sink i.e. ${Q_2}$ which can done using equation (i) as follows:
$
1 - \dfrac{{{Q_2}}}{{{Q_1}}} = 1 - \dfrac{{{T_2}}}{{{T_1}}}{\text{ }} \\
\Rightarrow \dfrac{{{Q_2}}}{{{Q_1}}} = \dfrac{{{T_2}}}{{{T_1}}} \\
\Rightarrow {Q_2} = \dfrac{{{T_2}}}{{{T_1}}} \times {Q_1} \\
$
Substituting the known values, we get
$
{Q_2} = \dfrac{{303}}{{996}} \times 9 \times {10^5} \\
= 2.737 \times {10^5}cal \\
$
Now we can calculate the work done by the Carnot engine using the equation (ii) as follows:
$W = {Q_1} - {Q_2}{\text{ }}$
Substituting the values, we get
$
W = 9 \times {10^5} - 2.737 \times {10^5} \\
\simeq 6.26 \times {10^5}cal \\
$
Hence, the correct answer is option C.
Note: The Carnot cycle is reversible in nature and its efficiency is the highest possible for an engine. The ideal temperature of the sink should be 0K to get maximum efficiency but it is not possible to practically attain this temperature due to which carnot engine is considered ideal in nature.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Which are the Top 10 Largest Countries of the World?
Difference Between Plant Cell and Animal Cell
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Give 10 examples for herbs , shrubs , climbers , creepers
Change the following sentences into negative and interrogative class 10 english CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE