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One reversible heat engine operates between 1600 K and T2 K, and smother reversible engine operates between T2 K and 400 K. If both engines have the same heat input and output, then the temperature T2 must be equal to
- a)1000 K
- b)1200 K
- c)1400 K
- d)800 K
Correct answer is option 'D'. Can you explain this answer?
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One reversible heat engine operates between 1600 K and T2 K, and smoth...
The given condition means equal efficiency of both engines , therefore
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One reversible heat engine operates between 1600 K and T2 K, and smoth...
If 2 rev engine operating bw thermal reservoir at T1,T2,T3 such that t1>t2>t3 and if the efficiency is same for both the engine then,T2=(T1*T3)^1/2
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One reversible heat engine operates between 1600 K and T2 K, and smoth...
To find the temperature T2, we can use the Carnot efficiency formula. The Carnot efficiency is the maximum efficiency that a heat engine can achieve when operating between two temperature reservoirs. It is given by the formula:
Efficiency = 1 - (T2 / T1)
where T1 is the higher temperature reservoir and T2 is the lower temperature reservoir.
In this case, we have two reversible heat engines operating between different temperatures. Let's analyze each engine separately:
1. Reversible Engine 1:
- Operates between 1600 K and T2 K
- Heat input (Q1) = Heat output (Q2)
2. Reversible Engine 2:
- Operates between T2 K and 400 K
- Heat input (Q2) = Heat output (Q3)
Since both engines have the same heat input and output, we can equate Q1 = Q3.
Now we can calculate the efficiency of each engine and equate them to find T2:
Efficiency1 = 1 - (T2 / 1600)
Efficiency2 = 1 - (400 / T2)
Since both engines have the same efficiency, we can equate the two equations:
1 - (T2 / 1600) = 1 - (400 / T2)
Simplifying the equation, we get:
T2 / 1600 = 400 / T2
Cross-multiplying, we get:
T2^2 = 1600 * 400
T2^2 = 640000
Taking the square root of both sides, we get:
T2 ≈ 800 K
Therefore, the temperature T2 must be approximately 800 K.
Efficiency = 1 - (T2 / T1)
where T1 is the higher temperature reservoir and T2 is the lower temperature reservoir.
In this case, we have two reversible heat engines operating between different temperatures. Let's analyze each engine separately:
1. Reversible Engine 1:
- Operates between 1600 K and T2 K
- Heat input (Q1) = Heat output (Q2)
2. Reversible Engine 2:
- Operates between T2 K and 400 K
- Heat input (Q2) = Heat output (Q3)
Since both engines have the same heat input and output, we can equate Q1 = Q3.
Now we can calculate the efficiency of each engine and equate them to find T2:
Efficiency1 = 1 - (T2 / 1600)
Efficiency2 = 1 - (400 / T2)
Since both engines have the same efficiency, we can equate the two equations:
1 - (T2 / 1600) = 1 - (400 / T2)
Simplifying the equation, we get:
T2 / 1600 = 400 / T2
Cross-multiplying, we get:
T2^2 = 1600 * 400
T2^2 = 640000
Taking the square root of both sides, we get:
T2 ≈ 800 K
Therefore, the temperature T2 must be approximately 800 K.
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One reversible heat engine operates between 1600 K and T2 K, and smother reversible engine operates between T2 K and 400 K. If both engines have the same heat input and output, then the temperature T2 must be equal toa)1000 K b) 1200 Kc) 1400 K d) 800 KCorrect answer is option 'D'. Can you explain this answer?
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