Question

A Carnot's engine working between 277°C and 27°C produces 100 kJ work. Calculate
the heat rejected from the engine.
(a) Q = 20 kJ
(b) Q = 40 kJ
(c) Q = 10 kJ
(d) Q = 30 kJ
Answer​

Answer 1

Explanation:

d is the correct answer hope it helps you with the proper answer

Answer 2

Given:

The Temperature of Source= 277°C ( $T_{1}$ )

The Temperature of the sink = 27°C ( $T_{2}$ )

Work done=100kJ (W)

To find:

The heat rejected from the engine.

Solution:

Let the heat rejected by the engine be $Q_{2}$

By using the formula we can calculate the efficiency of the Carnot engine:

η=1-$T_{2}/T_{1}$

η=1-$27/277$

η=0.9 or 90 percent

Now, use the formula to calculate the heat absorbed ($Q_{1}$) by the engine.

η=$\frac{W}{Q_{1} }$

$0.9=\frac{100}{Q_{1}}$

$Q_{1}=111.11kJ$

We know that work done $W=Q_{1}-Q_{2}$

Substitute the value of work done (W) and heat absorbed $Q_{1}$ to find the value of heat rejected $Q_{2}$ by the engine.

$100=111.11-Q_{2}$

$Q_{2}=111.11-100$

$Q_{2}=11.11$ ≈10kJ

On solving we get, the value of the heat rejected from the engine as 10kJ.

The heat rejected from the engine is 10kJ

Option(c) is the correct choice.