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A Carnot engine operates between the temperatures of 300 K and 600 K. If the engine produces 600 KJ of work, what is the entropy change during heat addition in kJ/K?


1. 1
2. 0.5
3. 1.5
4. 2

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Correct Answer - Option 4 : 2

Concept:

The thermal efficiency of the Carnot cycle is given by

\(\eta = 1 - \frac{{{T_L}}}{{{T_H}}}\)

efficiency is also given by

\(\eta = \frac{{{W_{out}}}}{{{Q_S}}}\)

The change of entropy is given by \(dS = \frac{{dQ}}{T}\) 

Calculation:

Given:

T1= 600K and T= 300K, W = 600 kJ

therefore

\(\eta = 1 - \frac{{{T_L}}}{{{T_H}}} = 1 - \frac{{300}}{{600}} = \frac{1}{2}\)

Now the \(\frac{1}{2} = \frac{{600}}{{{Q_S}}}\)

⇒ QS = 1200 kJ

Now, change in entropy of working fluid during heat addition process is given by:

\(dS = \frac{{dQ}}{T} = \frac{{1200}}{{600}} = 2\;kJ/K\)

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