Question

An ideal gas heat engine operates in the Carnot cycle between ${227^ \circ }C$ and${127^ \circ }C$. It absorbs $6 \times {10^4}cal$ of heat at higher temperature. Amount of heat converted into work, is
A) $1.2 \times {10^4}cal$
B) $2.4 \times {10^4}cal$
C) $6.0 \times {10^4}cal$
D) $4.8 \times {10^4}cal$

Answer 1

When a heat engine operates, it takes heat from the source, converts it into work and then sends remaining heat to the sink. In every heat engine, there are three main parts-

1. Source
2. Working substance
3. Sink
   Refrigerator is an example of a heat engine. Heat engines always take heat from the source. So, the temperature of the source is always greater than the sink.

Formula used- In this question, we can solve this by applying efficiency to the heat engine.
If a heat engine took heat from the source s input and converted it into work by working substance. Then efficiency is given by-
$\eta {\text{ = }}\dfrac{{{\text{work done by working substance}}}}{{{\text{heat absorbed by the source}}}}$
$\eta = \dfrac{W}{{{Q_1}}}$
$W$is the work done by the working substance i.e. $W = Q{}_1 - {Q_2}$
Where ${Q_1}$ and ${Q_2}$are the heat absorbed by the working substance and heat given to sink by working substance. So,
$\eta = \dfrac{{{Q_1} - {Q_2}}}{{{Q_1}}} \\\ \Rightarrow \eta = 1 - \dfrac{{{Q_2}}}{{{Q_1}}} \\\$
If $T{}_1$ and $T{}_2$ are the temperature of source and sink respectively.
$\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}$
$\eta$ is the efficiency of a heat engine.

Complete step-by-step answer:
In this question, a Carnot engine operates between ${227^ \circ }C$ and ${127^ \circ }C$. So, ${227^ \circ }C$ is the temperature of the source and ${127^ \circ }C$ is the temperature of the sink.
${T_1} = {227^ \circ }C \\\ \Rightarrow {T_1} = 227 + 273 \\\ \Rightarrow {T_1} = 500K \\\$
${T_2} = {127^ \circ }C \\\ \Rightarrow {T_1} = 127 + 273 \\\ \Rightarrow {T_1} = 400K \\\$
And also it absorbs $6 \times {10^4}cal$ from source. So, ${Q_1} = 6 \times {10^4}cal$
Now, efficiency of heat engine-
$\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}$
Substituting the values, we get-
$\eta = 1 - \dfrac{{400}}{{500}} \\\ \Rightarrow \eta = 1 - \dfrac{4}{5} \\\ \Rightarrow \eta = \dfrac{{5 - 4}}{5} \\\ \Rightarrow \eta = \dfrac{1}{5} \\\$
Also efficiency is known as
$\eta = \dfrac{W}{{{Q_1}}}$
Substituting the values, we get-

$$\Rightarrow \dfrac{1}{5} = \dfrac{W}{{6 \times {{10}^4}}} \\\ \Rightarrow W = \dfrac{{6 \times {{10}^4}}}{5} \\\ \Rightarrow W = 1.2 \times {10^4}cal \\\$$

Hence, option A is correct.

Note: - It is clear that efficiency of any heat engine depends on temperature of source and sink. And the efficiency of any heat engine is always less than $1$. In practice there is no possible heat engine with efficiency $1$.