Question

An engine has an efficiency of $\dfrac{1}{3}$. The amount of work this engine can perform per kilocalorie of heat input is
(A) $1400\,cal$
(B) $700\,cal$
(C) $700\,J$
 (D) $1400\,J$

Answer 1

Hint:Use the formula of the efficiency of the heat engine. Substitute the values of the efficiency and the heat input in the formula, to obtain the work done by the engine. Use the value of the heat input in both calories and joules to know the correct answer, since the option contains both the units.

Useful formula:
The formula for the efficiency is given by
$\eta = \dfrac{w}{h}$
Where $\eta $ is the efficiency of the engine, $w$ is the work done by it and $h$ is the heat input into the engine.

Complete step by step solution:
It is given that the
Efficiency of the heat engine, $\eta = \dfrac{1}{3}$
The heat input of the engine, $h = 1\,kcal$
If we convert the above value of heat input into the calories or joules. The value obtained is $1000\,cal$ when it was converted into calories and $4.2 \times {10^{3\,}}\,J$, when it is written in joules. Since one calorie is equal to the $4.2$ kilo joules.
Use the formula of efficiency.
$\eta = \dfrac{w}{h}$
Substitute the values of the efficiency of the engine and the heat input of it.
$\dfrac{1}{3} = \dfrac{w}{1}$
Changing the value of the heat input in joules.
$\dfrac{1}{3} = \dfrac{w}{{4.2 \times 1000}}$
By further simplification to obtain the value of the work done.
$w = \dfrac{{4200}}{3}$
By performing division in the above step.
$w = 1400\,J$
Hence work done of the engine is obtained as $1400\,J$.

Thus the option (D) is correct.

Note:But if we write the heat input in the calories, the value of the heat input is $1000\,cal$. Hence the work done obtained from that is $\dfrac{{1000}}{3}$, which is equal to $333.3\,cal$. This answer is not in the option. And so the work done in the engine is obtained as the joules.