Question

A carnot engine absorbs 1000J of heat energy from a reservoir at ${{127}^{\circ }}C$ and rejects 600J of heat energy during each cycle. Calculate
A. efficiency of the engine,
B. temperature of sink,
C. amount of useful work done per cycle.

Answer 1

At first we need to write all the values that are given in the question, then for the first part that is efficiency there is a formula for efficiency of carnot engine write that and replace the values to get the result. Now there is a relation between heat absorbed or rejected and initial and final temperature of the engine with that formula we will get the temperature of the sink. Now for useful work done there is also a formula just write it and then replace with the values to get the required result.

Formula used: $\eta =\left( 1-\dfrac{{{Q}_{1}}}{{{Q}_{2}}} \right)\times 100\%$
$\dfrac{{{Q}_{2}}}{{{Q}_{1}}}=\dfrac{{{T}_{2}}}{{{T}_{1}}}$
$W={{Q}_{1}}-{{Q}_{2}}$

Complete step by step answer:
So, at first from the question we will gather all the possible values that are,
We can see that the heat absorbed is ${{Q}_{1}}=1000J$
And the heat that is getting rejected is ${{Q}_{2}}=600J$
And the initial heat is ${{127}^{\circ }}C$ that is ${{T}_{1}}=\left( 127+273 \right)K=400K$
Now, as the first part of the question says, we have to find the efficiency of the reservoir.
We know that the formula for efficiency is,
$\eta =\left( 1-\dfrac{{{Q}_{1}}}{{{Q}_{2}}} \right)\times 100\%$
So, on putting all the given values
$\eta =\left( 1-\dfrac{600}{1000} \right)\times 100\%$,
Which on solving,
$\eta =40\%$, so we can say that the engine has an efficiency of 40%.
Now, as the second part of the question says we have to find the temperature of the sink,
We know that for carnot engine, $\dfrac{{{Q}_{2}}}{{{Q}_{1}}}=\dfrac{{{T}_{2}}}{{{T}_{1}}}$
Or,
${{T}_{2}}=\dfrac{{{Q}_{2}}}{{{Q}_{1}}}\times {{T}_{1}}$
So, on putting all the values,
$\Rightarrow {{T}_{2}}=\dfrac{600}{1000}\times 400$
On solving,
$\Rightarrow {{T}_{2}}=240K={{\left( 240-273 \right)}^{\circ }}C=-{{33}^{\circ }}C$
So, the temperature of the sink is, ${{-33}^{\circ }}C$.
Now we are asked to find the useful work done, so we know that for carnot engine useful work done will be,
$W={{Q}_{1}}-{{Q}_{2}}$
On placing the values,
$W=1000-600J$
$W=400J$.
Therefore, the engine has an efficiency of 40%, the temperature of the sink is ${{-33}^{\circ }}C$ and the amount of useful work done per cycle is $W=400J$.

Note: In the formula $\eta =\left( 1-\dfrac{{{Q}_{1}}}{{{Q}_{2}}} \right)\times 100\%$, ${{Q}_{1}}$ is the heat absorbed by the engine and ${{Q}_{2}}$is the heat that the engine rejected, it is multiplied by 100 to get the result in a percentage form. In the next equation that is $\dfrac{{{Q}_{2}}}{{{Q}_{1}}}=\dfrac{{{T}_{2}}}{{{T}_{1}}}$, ${{T}_{2}}$ is the final temperature and${{T}_{1}}$ is the initial temperature.