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An ideal gas heat engine operates in a Carnot cycle between 227ºC and 127ºC. It absorbs 6 kcal at the higher temperature. The amount of heat (in kcal) converted into work is equal to
  • a)
    1.2
  • b)
    4.8
  • c)
    3.5
  • d)
    1.6
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
An ideal gas heat engine operates in a Carnot cycle between 227ºC and...
Efficiency = T1 - T2 / T1
T1 = 227 + 273 = 500 K
T2= 127 + 273 = 400K
Hence , V = (η) x H = ⅕ x 6 = 1.2 kcal
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Community Answer
An ideal gas heat engine operates in a Carnot cycle between 227ºC and...
To solve this problem, we can use the Carnot efficiency formula, which is given by:

η = 1 - (Tc/Th)

where η is the efficiency, Tc is the absolute temperature at the cold reservoir, and Th is the absolute temperature at the hot reservoir.

Given that the temperatures are 227ºC and 127ºC, we need to convert them to absolute temperature by adding 273 to each:

Th = 227ºC + 273 = 500 K
Tc = 127ºC + 273 = 400 K

Now we can calculate the efficiency:

η = 1 - (400/500) = 1 - 0.8 = 0.2

The efficiency of the Carnot cycle is 0.2, which means that 20% of the heat energy absorbed is converted into work.

The problem states that the engine absorbs 6 kcal at the higher temperature. Since only 20% of this heat is converted into work, we can calculate the amount of heat converted into work:

Work = Efficiency * Heat absorbed
Work = 0.2 * 6 kcal = 1.2 kcal

Therefore, the amount of heat converted into work is 1.2 kcal, which corresponds to option A.
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An ideal gas heat engine operates in a Carnot cycle between 227ºC and 127ºC. It absorbs 6 kcal at the higher temperature. The amount of heat (in kcal) converted into work is equal toa)1.2b)4.8c)3.5d)1.6Correct answer is option 'A'. Can you explain this answer?
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