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A reversible heat engine receives 2 kJ of heat from a reservoir at 1000 K and a certain amount of heat from a reservoir at 800 K.It rejects 1 kJ of heat to a reservoir at 400 K.
The net work output (in kJ) of the cycle is
[2014, Set-1]
  • a)
    0.8
  • b)
    1.0
  • c)
    1.4
  • d)
    2.0
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A reversible heat engine receives 2 kJ of heat from a reservoir at 100...

We know that for reversible heat engine, change in entropy is always zero
That is ΔS = 0

Q2 = 0.4 kJ
WNet = (Q1 + Q2) – Q= (2 + 0.4) – 1 = 1.4 kJ.
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Most Upvoted Answer
A reversible heat engine receives 2 kJ of heat from a reservoir at 100...
Given:
Heat received from reservoir at 1000 K, Q1 = 2 kJ
Heat received from reservoir at 800 K, Q2 = x
Heat rejected to reservoir at 400 K, Q3 = 1 kJ

We know that the efficiency of a reversible heat engine is given by:
Efficiency = 1 - (Q3/Q1)

Using this, we can find the value of Q2:
Efficiency = 1 - (1/2)
Efficiency = 0.5

0.5 = 1 - (Q3/Q2 + Q1)
0.5 = 1 - (1/Q2 + 2)
1/Q2 = 2.5
Q2 = 0.4 kJ

Now, we can find the net work output of the cycle:
Net work output = Q1 - Q3
Net work output = 2 - 1
Net work output = 1 kJ

Therefore, option (c) is the correct answer.

Summary:
Efficiency = 1 - (Q3/Q1)
Using efficiency, we can find Q2
Net work output = Q1 - Q3
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A reversible heat engine receives 2 kJ of heat from a reservoir at 1000 K and a certain amount of heat from a reservoir at 800 K.It rejects 1 kJ of heat to a reservoir at 400 K.The net work output (in kJ) of the cycle is[2014,Set-1]a)0.8b)1.0c)1.4d)2.0Correct answer is option 'C'. Can you explain this answer?
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