Question

An ideal engine operates by taking in steam from a boiler at a temperature of ${327^0}C$ and rejecting heat to the sink at a temperature of ${27^0}C$ . The engine runs at $500\,rpm$ and heat taken is $600\,kcal$ in each revolution. Calculate
A. The Carnot efficiency of the engine
B. The work done in each cycle
C. The heat rejected in each revolution
D. The power output of the engine

Answer 1

In thermal physics, Efficiency of an engine is the ratio of net output power to the net input power of an engine and efficiency of an engine is always expressed in percentage. The revolution per minute is the unit of measuring the frequency of an engine.

Complete step by step answer:
(a) Given that, the temperature of source is ${T_1} = {327^0}C = 600K$
The temperature of sink is ${T_2} = {27^0}C = 300K$
Now, the efficiency of an engine is given as
$\text{efficiency} = 1 - \dfrac{{{T_2}}}{{{T_1}}}$
On putting the values of temperature we get,
$\text{efficiency}= 1 - \dfrac{{300}}{{600}}$
Converting it in to percentage
$\therefore \text{efficiency}= 50\%$

Hence, the Carnot efficiency of engine is $50\%$

(b) Now, efficiency of the engine can also be written in form of work done and heat as
$\text{efficiency} = \dfrac{W}{{{Q_1}}}$ Where, ${Q_1} = 600\,kcal$ as given in the question
Putting the value of heat we have,
$W = \text{efficiency} \times 600$
$\Rightarrow W = 0.5 \times 600$
$\therefore W = 300\,kcal$

Hence, the work done in each cycle by the engine is $300\,kcal$.

(c) Now, to find rejected heat, we can simple subtract the given heat from work done so,
${Q_2} = W - {Q_1}$
Which we get,
${Q_2} = 600 - 300$
$\therefore {Q_2} = 300kcal$

Hence, the heat rejected in each cycle is $300\,kcal$.

(d) Now, time period $T = \dfrac{1}{f}$ and frequency is given $500\,rpm$
Power of engine is $\text{Power} = W \times \dfrac{1}{T}$
We need to convert time in seconds so, we have
Time for one cycle is $T = \dfrac{{60}}{{500}}$
$T = 0.12\sec$
$\Rightarrow \text{Power} = \dfrac{{300}}{{0.12}}$
$\therefore \text{Power}= 2500\,Kcal{\sec ^{ - 1}}$

Hence, the output power of the engine is $2500\,Kcal\,{\sec ^{ - 1}}$.

Note: We must remember some basic conversions used in the solution like, The conversion of $^0C$ scale of temperature into kelvin scale is given by $kelvin{ = ^0}C + 273$ and Revolution per minute (rpm) is the one revolution per minute which can be written as $\dfrac{1}{{60}}(revolution)$ in seconds.