An ideal gas heat engine operates in a Carnot Cycle between temperatures of 266 and 115°C. It absorbs 4250 cal per cycle at 266°C. What is its efficiency as a percentage?
Thot = 266 + 273 = 539
Tcold = 115 + 273 = 388
That is all you need for Carnot efficiency
eff = 100 (1 - Tc/Th) = 100(1-388/539)
=28 percent
To find the efficiency of the Carnot Cycle heat engine, we need to first calculate the heat absorbed and then the heat rejected.
Step 1: Calculate the heat absorbed (Qh):
Given that the heat absorbed by the engine is 4250 calories per cycle, we have:
Qh = 4250 cal
Step 2: Calculate the temperature difference between the two reservoirs (ΔT):
The temperature difference between the hot reservoir (Th) and the cold reservoir (Tc) is given by:
ΔT = Th - Tc
Given that Th = 266°C and Tc = 115°C, we can convert these temperatures to Kelvin:
Th = 266 + 273 = 539 K
Tc = 115 + 273 = 388 K
ΔT = 539 K - 388 K = 151 K
Step 3: Calculate the heat rejected (Qc):
The heat rejected by the engine can be calculated using the formula for the efficiency of a Carnot Cycle:
Efficiency = 1 - (Qc / Qh)
Since the Carnot Cycle is a reversible process, the efficiency can also be expressed as:
Efficiency = 1 - (Tc / Th)
Substituting the values for Tc and Th:
Efficiency = 1 - (388 K / 539 K)
Step 4: Calculate the efficiency as a percentage:
To express the efficiency as a percentage, we multiply it by 100:
Efficiency (in %) = (1 - (388 K / 539 K)) * 100
Calculating this expression will give us the efficiency of the Carnot Cycle heat engine as a percentage.
Please note that in this particular case, we are not given the heat rejected by the engine (Qc), so we are unable to provide the exact value of the efficiency.