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In a thermodynamic cycle consisting of four processes, the heat and work are as follows:
Q : 30, – 10, – 20, 5
W: 3, 10, – 8, 0
The thermal efficiency of the cycle will be:
a) 
Zero
  • b) 
    7.15 %
  • c) 
    14.33 %
  • d) 
    28.6 %
Correct answer is option 'C'. Can you explain this answer?
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Thermodynamic Cycle Analysis:

Given, Q1 = 30, Q2 = -10, Q3 = -20, Q4 = 5
W1 = 3, W2 = 10, W3 = -8, W4 = 0

The thermal efficiency is given by the ratio of net work output to the heat input.

Net Work Output:
Wnet = W1 + W2 + W3 + W4 = 3 + 10 - 8 + 0 = 5 J

Heat Input:
Qin = Q1 + Q2 + Q3 + Q4 = 30 - 10 - 20 + 5 = - 5 J

As heat input is negative, we can write,
|Qin| = 5 J

Thermal Efficiency:
ηth = Wnet/|Qin| = 5/5 = 1

Therefore, the thermal efficiency of the cycle is 1 or 100%.

Answer: Option C (14.33%) is incorrect.
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