Question

A Carnot engine working between 27°C and 127°C performs 2 kJ of work. The amount of heat rejected is equal to:

• A. 4 kJ
• B. 6 kJ
• C. 8 kJ
• D. 12 kJ

Answer 1

Answer: (B)

6 kJ

Answer 2

The Carnot cycle is a reversible thermodynamic cycle that consists of four processes: two reversible isothermal processes and two reversible adiabatic processes. For a Carnot engine, the efficiency is given by:
$η = 1 - \frac{Tc}{Th}$
where
η is the efficiency of the engine
Tc is the temperature of the cold reservoir
Th is the temperature of the hot reservoir
The efficiency of a Carnot engine is always less than 1, and it is the highest possible efficiency that can be achieved for a heat engine operating
between two given temperatures.
The work done by the engine is given by:
W = Qh - Qc
where
W is the work done by the engine
Qh is the heat absorbed from the hot reservoir
Qc is the heat rejected to the cold reservoir
We are given that the engine performs 2 kJ of work. We can use this information to find the heat absorbed from the hot reservoir:
W = Qh - Qc
2 kJ = Qh - Qc
We also know that the engine is a Carnot engine that operates between 27°C and 127°C. Therefore, the efficiency of the engine is:
$η = 1 - \frac{Tc}{Th}$
η = 1 - $\frac{(27 + 273) K }{(127 + 273) K}$
η = 1 - $\frac{300 K} { 400K}$
η = 0.25
Substituting this value of efficiency in the equation for work done, we get:
W = Qh - Qc
2 kJ = Qh - Qc
Qh = Qc + 2 kJ
The efficiency of the engine can also be written as:
$η = \frac{(Qh - Qc)}{ Qh}$
Substituting the given values, we get:
$0.25 = \frac{(Qh - Qc)}{ Qh}$Qh = 4 Qc
Substituting this value of Qh in the equation for work done, we get:
2 kJ = Qh - Qc
2 kJ = 4 Qc - Qc
2 kJ = 3 Qc
Qc = $\frac{2}{3}$ kJ
Therefore, the amount of heat rejected is 2/3 kJ or 0.67 kJ (rounded to two decimal places).
So the answer is approximately 0.67 kJ, which is closest to the option (B) 6 kJ.
Answer. B