Question

A Carnot heat engine works between 300K and 600K. If it consumes 100J in each cycle, what is the heat rejected in each cycle? What is the work done?

Answer 1

Hint: To find the heat rejected per cycle and the work done, find the efficiency of the cycle first. Use the formula for efficiency of an engine, substitute the values and calculate the efficiency of the engine. Then, to calculate the heat rejected per cycle, calculate the work done. To find the work done, use the formula giving relation between work done, efficiency and heat absorbed. Substitute the values and obtain the amount of work done. Now, subtract the heat absorbed from the work done. This will give the amount of heat rejected per cycle.

Complete answer:
Given: Temperature of source (${T}_{1})= 600K$
            Temperature of sink (${T}_{2})= 300K$
            Heat absorbed (${Q}_{1}$) = 100J
Efficiency of a Carnot engine is given by,
$\eta = 1- \dfrac {{T}_{2}}{{T}_{1}}$
Substituting the values in above equation we get,
$\eta= 1- \dfrac {300}{600}$
$\Rightarrow \eta= 1-0.5$
$\Rightarrow \eta=0.5$
$\Rightarrow \eta=50\%$
Let W be the work done
We know, relation between work done and efficiency is given by,
$\eta = \dfrac {W}{{Q}_{1}}$
Substituting values in above equation we get,
$0.5 = \dfrac {W}{100}$
$\Rightarrow W= 0.5 \times 100$
$\Rightarrow W= 50J$
Let the heat rejected be ${Q}_{2}$
$W ={Q}_{1}-{Q}_{2}$
Substituting the values in above equation we get,
$50 =100-{Q}_{2}$
$\Rightarrow {Q}_{2} = 100-50$
$\Rightarrow {Q}_{2} = 50J$
Hence, the heat rejected in each cycle is 50J. Work done is also 50J.

Note:
Efficiency of the Carnot engine can be increased by decreasing the temperature of the sink. As the unit of temperatures is already in Kelvin, we don’t have to convert it. Don’t get confused between the temperature of source and temperature of sink. Temperature of the source is always greater than the temperature of the sink. It is not possible to attain $100\%$ efficiency for a Carnot engine. To attain such an efficiency, the input heat should be completely converted to work. Converting heat completely into work is not possible.