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The efficiency of a carnot engine is 50%. The temperature of the hot reservoir is kept constant. By what amount should the temperature of the cold reservoir be decreased so that efficiency becomes 60%.
- a)1K
- b)2K
- c)30K
- d)83K
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
The efficiency of a carnot engine is 50%. The temperature of the hot r...
0.5 = 1 – TC/TH.
And 0.6 = 1 – (TC + x)/TH.
From the first equation we get: 5/10 = TC/TH.
From the second equation we get: 4/10 = (TC + x)/TH.
Now, we see that the value of x must be 1K.
And 0.6 = 1 – (TC + x)/TH.
From the first equation we get: 5/10 = TC/TH.
From the second equation we get: 4/10 = (TC + x)/TH.
Now, we see that the value of x must be 1K.
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The efficiency of a carnot engine is 50%. The temperature of the hot r...
To solve this problem, we need to use the formula for the efficiency of a Carnot engine:
Efficiency = 1 - (Tc/Th)
Where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir.
Let's assume the initial temperature of the hot reservoir is Th and the efficiency is 50%. According to the given equation, we have:
0.5 = 1 - (Tc/Th)
Rearranging the equation, we get:
Tc/Th = 1 - 0.5
Tc/Th = 0.5
Now, let's assume the new temperature of the cold reservoir is Tc' and the efficiency is 60%. We can write the equation as:
0.6 = 1 - (Tc'/Th)
Rearranging the equation, we get:
Tc'/Th = 1 - 0.6
Tc'/Th = 0.4
To find the change in temperature, we subtract the two equations:
(Tc'/Th) - (Tc/Th) = 0.4 - 0.5
Simplifying, we get:
Tc'/Th - Tc/Th = -0.1
Since Th is constant, we can cancel it out:
Tc' - Tc = -0.1Th
Since we are looking for the change in temperature, we can rearrange the equation as:
Tc' = Tc - 0.1Th
Now, we substitute the value of Tc/Th from the initial equation:
Tc' = Tc - 0.1(0.5)
Simplifying, we get:
Tc' = Tc - 0.05
Since we want to increase the efficiency, the temperature of the cold reservoir should be decreased. Therefore, the change in temperature is negative.
To increase the efficiency from 50% to 60%, the temperature of the cold reservoir should be decreased by 0.05 units.
The correct answer is option A) 1K.
Efficiency = 1 - (Tc/Th)
Where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir.
Let's assume the initial temperature of the hot reservoir is Th and the efficiency is 50%. According to the given equation, we have:
0.5 = 1 - (Tc/Th)
Rearranging the equation, we get:
Tc/Th = 1 - 0.5
Tc/Th = 0.5
Now, let's assume the new temperature of the cold reservoir is Tc' and the efficiency is 60%. We can write the equation as:
0.6 = 1 - (Tc'/Th)
Rearranging the equation, we get:
Tc'/Th = 1 - 0.6
Tc'/Th = 0.4
To find the change in temperature, we subtract the two equations:
(Tc'/Th) - (Tc/Th) = 0.4 - 0.5
Simplifying, we get:
Tc'/Th - Tc/Th = -0.1
Since Th is constant, we can cancel it out:
Tc' - Tc = -0.1Th
Since we are looking for the change in temperature, we can rearrange the equation as:
Tc' = Tc - 0.1Th
Now, we substitute the value of Tc/Th from the initial equation:
Tc' = Tc - 0.1(0.5)
Simplifying, we get:
Tc' = Tc - 0.05
Since we want to increase the efficiency, the temperature of the cold reservoir should be decreased. Therefore, the change in temperature is negative.
To increase the efficiency from 50% to 60%, the temperature of the cold reservoir should be decreased by 0.05 units.
The correct answer is option A) 1K.
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The efficiency of a carnot engine is 50%. The temperature of the hot reservoir is kept constant. By what amount should the temperature of the cold reservoir be decreased so that efficiency becomes 60%.a)1Kb)2Kc)30Kd)83KCorrect answer is option 'A'. Can you explain this answer?
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