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. The efficiency of engine and temperature of sink will be respectively
A. 20% and$-43{}^\circ C$
B. 40% and $-33{}^\circ C$
C. 50% and$-20{}^\circ C$
D. 70% and$-10{}^\circ C$

Answer 1

As a first step you could recall the expression for efficiency of Carnot engine. We are given the heat energy absorbed, temperature of the reservoir and the heat energy rejected during each cycle, all ready to be substituted. Substituting these values will get you the efficiency of engine and temperature of sink from the expression for efficiency of Carnot engine.
Formula used:
Expression for efficiency of a Carnot engine,
$\eta =1-\dfrac{{{Q}_{2}}}{{{Q}_{1}}}=1-\dfrac{{{T}_{2}}}{{{T}_{1}}}$

Complete answer:
In the question, we are given a Carnot engine that absorbs 1000J of heat energy from a reservoir that is kept at$127{}^\circ C$ , that is,
Let, Heat absorbed from the reservoir be given by, ${{Q}_{1}}$ then,
${{Q}_{1}}=1000J$
And let ${{T}_{1}}$ be the temperature of the reservoir, then,
${{T}_{1}}=127{}^\circ C=400K$
Also the given Carnot engine rejects 600J of heat energy during each cycle, that is, if the heat rejected to the sink from the Carnot engine is given by, ${{Q}_{2}}$ , then,
${{Q}_{2}}=600J$
With these given values, we are asked to find the efficiency of the engine as well as the temperature of the sink.
Let us recall the expression for the Carnot engine.
If we represent the net efficiency of Carnot engine by$\eta$ , and then,
$\eta =\dfrac{W}{{{Q}_{1}}}$
Where, W is the net work done by the gas given by,
$W={{Q}_{1}}-{{Q}_{2}}$
$\Rightarrow \eta =\dfrac{{{Q}_{1}}-{{Q}_{2}}}{{{Q}_{1}}}=1-\dfrac{{{Q}_{2}}}{{{Q}_{1}}}$ …………………………… (1)
Substituting the given values from the question,
$\Rightarrow \eta =\dfrac{1000-600}{1000}$
$\Rightarrow \eta =0.4$
So the percentage efficiency of the Carnot engine = 40% ……………………….. (2)
The efficiency of the Carnot engine has another expression which is given by,
$\eta =1-\dfrac{{{T}_{2}}}{{{T}_{1}}}$ …………………… (3)
Comparing (1) and (3), we get the universal relation for a Carnot cycle that is independent of the nature of the system,
$\dfrac{{{Q}_{1}}}{{{Q}_{2}}}=\dfrac{{{T}_{1}}}{{{T}_{2}}}$
Rearranging we get the temperature of the sink ${{T}_{2}}$ as,
${{T}_{2}}=\dfrac{{{T}_{1}}\times {{Q}_{2}}}{{{Q}_{1}}}$
$\Rightarrow {{T}_{2}}=\dfrac{400K\times 600J}{1000J}$
$\Rightarrow {{T}_{2}}=240K$
$\Rightarrow {{T}_{2}}=240-273=-33{}^\circ C$
Therefore, the temperature of the sink is $-33{}^\circ C$ ………………………. (4)

So, the correct answer is “Option B”.

Note:
Don’t forget to substitute the given temperature in its SI unit in the expression for Efficiency of Carnot engine. Since the options are given in degrees Celsius you convert the temperature of the sink back to Celsius from Kelvin. Also, convert the efficiency into percentage value as the options contain percentage efficiency.