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Two reversible engine operate between thermal reservoirs at 1200 K, T2K and 300 K such that 1st engine receives heat from 1200 K reservoir and rejects heat to thermal reservoir at T2K, while the 2nd engine receives heat from thermal reservoir at T2K and rejects heat to the thermal reservoir at 300 K. The efficiency of both the engines is equal.
What is the value of temperature T2?
- a)400 K
- b)500 K
- c)600 K
- d)700 K
Correct answer is option 'C'. Can you explain this answer?
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Two reversible engine operate between thermal reservoirs at 1200 K, T2...
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Two reversible engine operate between thermal reservoirs at 1200 K, T2...
Given information:
- Two reversible engines operate between thermal reservoirs at 1200 K, T2K, and 300 K.
- The first engine receives heat from the 1200 K reservoir and rejects heat to the thermal reservoir at T2K.
- The second engine receives heat from the thermal reservoir at T2K and rejects heat to the thermal reservoir at 300 K.
- The efficiency of both engines is equal.
To find the value of temperature T2, we can use the Carnot efficiency formula:
Efficiency = 1 - (Tc/Th)
Where:
- Efficiency is the ratio of useful work output to the heat input.
- Tc is the temperature of the cold reservoir.
- Th is the temperature of the hot reservoir.
Let's calculate the efficiencies of both engines separately:
First Engine:
- Hot reservoir temperature (Th) = 1200 K
- Cold reservoir temperature (Tc) = T2K
Efficiency of the first engine = 1 - (Tc/Th)
= 1 - (T2/1200)
Second Engine:
- Hot reservoir temperature (Th) = T2K
- Cold reservoir temperature (Tc) = 300 K
Efficiency of the second engine = 1 - (Tc/Th)
= 1 - (300/T2)
Since the efficiencies of both engines are equal, we can equate the two expressions:
1 - (T2/1200) = 1 - (300/T2)
Simplifying the equation:
T2/1200 = 300/T2
Cross-multiplying:
T2^2 = 1200 * 300
T2^2 = 360000
Taking the square root of both sides:
T2 = √360000
T2 ≈ 600 K
Therefore, the value of temperature T2 is 600 K. Hence, option C is the correct answer.
- Two reversible engines operate between thermal reservoirs at 1200 K, T2K, and 300 K.
- The first engine receives heat from the 1200 K reservoir and rejects heat to the thermal reservoir at T2K.
- The second engine receives heat from the thermal reservoir at T2K and rejects heat to the thermal reservoir at 300 K.
- The efficiency of both engines is equal.
To find the value of temperature T2, we can use the Carnot efficiency formula:
Efficiency = 1 - (Tc/Th)
Where:
- Efficiency is the ratio of useful work output to the heat input.
- Tc is the temperature of the cold reservoir.
- Th is the temperature of the hot reservoir.
Let's calculate the efficiencies of both engines separately:
First Engine:
- Hot reservoir temperature (Th) = 1200 K
- Cold reservoir temperature (Tc) = T2K
Efficiency of the first engine = 1 - (Tc/Th)
= 1 - (T2/1200)
Second Engine:
- Hot reservoir temperature (Th) = T2K
- Cold reservoir temperature (Tc) = 300 K
Efficiency of the second engine = 1 - (Tc/Th)
= 1 - (300/T2)
Since the efficiencies of both engines are equal, we can equate the two expressions:
1 - (T2/1200) = 1 - (300/T2)
Simplifying the equation:
T2/1200 = 300/T2
Cross-multiplying:
T2^2 = 1200 * 300
T2^2 = 360000
Taking the square root of both sides:
T2 = √360000
T2 ≈ 600 K
Therefore, the value of temperature T2 is 600 K. Hence, option C is the correct answer.
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Two reversible engine operate between thermal reservoirs at 1200 K, T2K and 300 K such that 1st engine receives heat from 1200 K reservoir and rejects heat to thermal reservoir at T2K, while the 2nd engine receives heat from thermalreservoir at T2K and rejects heat to the thermal reservoir at 300 K. The efficiency of both the engines is equal.What is the value of temperature T2?a)400 Kb)500 Kc)600 Kd)700 KCorrect answer is option 'C'. Can you explain this answer?
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