Question

A Carnot engine takes 900 kcal of heat from a reservoir at ${723}^{\circ }C$ and exhausts it to a sink at ${30}^{\circ }C$. The work done by the engine is:
$$\begin{gathered} \text{A}\text{. 2}\text{.73}\times \text{1}{\text{0}}^6\text{ Cal} \\ \text{B}\text{. 3}\text{.73}\times \text{1}{\text{0}}^6\text{ Cal} \\ \text{C}\text{. 6}\text{.27}\times \text{1}{\text{0}}^6\text{ Cal} \\ \text{D}\text{. 5}\text{.73}\times \text{1}{\text{0}}^6\text{ Cal} \end{gathered}$$

Answer 1

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
$\eta =1-{\displaystyle \frac{Q_2}{Q_1}}=1-{\displaystyle \frac{T_2}{T_1}}...\text{(i)}$
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
$W=Q_1-Q_2...\text{(ii)}$

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:
$$\begin{gathered} 1-{\displaystyle \frac{Q_2}{Q_1}}=1-{\displaystyle \frac{T_2}{T_1}} \\ \Rightarrow {\displaystyle \frac{Q_2}{Q_1}}={\displaystyle \frac{T_2}{T_1}} \\ \Rightarrow Q_2={\displaystyle \frac{T_2}{T_1}}\times Q_1 \end{gathered}$$
Substituting the known values, we get
$$\begin{gathered} Q_2={\displaystyle \frac{303}{996}}\times 9\times {10}^5 \\ =2.737\times {10}^{5}cal \end{gathered}$$
Now we can calculate the work done by the Carnot engine using the equation (ii) as follows:
$W=Q_1-Q_2$
Substituting the values, we get
$$\begin{gathered} W=9\times {10}^5-2.737\times {10}^5 \\ \simeq 6.26\times {10}^{5}cal \end{gathered}$$
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.