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A heat engine working on Cannot cycle absorbs heat from three reservoirs at 1000 K, 800 K and 600 K. The engine does 10 kW of net work and rejects 400 kJ/min of heat to a heat sink at 300 K. If heat supplied by the reservoir at 1000 K is 50% of the heat supplied by reservoir at 600 K, the quantity of heat exchanged with the reservoir at 800 K will be _____ (in kW).
  • a)
    4.8 kW
  • b)
    9.5 kW
  • c)
    2.4 kW
  • d)
    7.2 kW
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
A heat engine working on Cannot cycle absorbs heat from three reservoi...

 
W = 10 kW, Q4 = 400 kJ/min = 6.67 kW
Q1 = 0.5 Q3
Energy balance:

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Most Upvoted Answer
A heat engine working on Cannot cycle absorbs heat from three reservoi...
To solve this problem, we need to apply the principles of thermodynamics and the laws of heat transfer. Let's break down the given information and solve step by step.

Given:
- The heat engine operates on a Cannot cycle.
- Heat is absorbed from three reservoirs at temperatures of 1000 K, 800 K, and 600 K.
- The net work done by the engine is 10 kW.
- Heat is rejected to a heat sink at 300 K at a rate of 400 kJ/min.
- The heat supplied by the 1000 K reservoir is 50% of the heat supplied by the 600 K reservoir.

Step 1: Calculate the heat supplied by the 1000 K reservoir.
Let's assume the heat supplied by the 600 K reservoir is Q6. Therefore, the heat supplied by the 1000 K reservoir is 0.5*Q6.

Step 2: Calculate the heat supplied by the 800 K reservoir.
Since the heat engine operates on a Cannot cycle, the heat supplied by the 800 K reservoir (Q8) can be calculated using the heat balance equation:
Q8 = Q6 + Q6 - Q10 - Q12

Step 3: Calculate the heat rejected to the heat sink.
The heat rejected to the heat sink (Qc) is given as 400 kJ/min.

Step 4: Calculate the net work done by the engine.
The net work done by the engine (Wnet) is given as 10 kW.

Step 5: Calculate the heat supplied by the 600 K reservoir.
Since the heat supplied by the 600 K reservoir is Q6, we can calculate it by rearranging the heat balance equation:
Q6 = Q10 + Q12 + Qc

Step 6: Substitute the values and solve for Q6.
Using the known values:
Qc = 400 kJ/min = 400/60 = 6.67 kW
Q10 = 0.5*Q6 = 0.5*(Q10 + Q12 + 6.67)
Q12 = 0.5*Q6 = 0.5*(Q10 + Q12 + 6.67)
Wnet = Q10 + Q12 - Qc

By solving these equations, we find that Q6 = 12.5 kW.

Step 7: Substitute the values and solve for Q8.
Q8 = Q6 + Q6 - Q10 - Q12
Q8 = 12.5 + 12.5 - 0.5*(12.5 + 12.5 + 6.67)
Q8 = 9.375 kW

Therefore, the quantity of heat exchanged with the reservoir at 800 K is 9.375 kW, which matches option (B).
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A heat engine working on Cannot cycle absorbs heat from three reservoirs at 1000 K, 800 K and 600 K. The engine does 10 kW of net work and rejects 400 kJ/min of heat to a heat sink at 300 K. If heat supplied by the reservoir at 1000 K is 50% of the heat supplied by reservoir at 600 K, the quantity of heat exchanged with the reservoir at 800 K will be _____ (in kW).a)4.8 kWb)9.5 kWc)2.4 kWd)7.2 kWCorrect answer is option 'B'. Can you explain this answer?
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