Verified answer

In an ideal cyclical engine, efficiency is defined as e = w / Qh (work / heat flow from hot reserviour). Based on the 2nd law of thermodynamics, in a closed system with each reserviour at constant temperature, Qh / Th + Qc / Tc = 0. Work is also the difference between Qh and Qc. Qh - Qc = w.

So:

e = 1 - Qc/Qh = 1 - Tc/Th

where Tc is 270 K and Th is 2700 K.

Not at all clear what you are asking here. Ideally we'd like 100% efficiency. That is, we'd like to get out as useful work all the energy put into a system.

But if you're asking the theoretical maximum efficiency from a Carnot cycle; then that's [(TH - TC)/TH] X 100% = ((2700-270)/2700] X 100% efficiency...you can do the arithmetic. [See source.]

The Carnot cycle is, through definition, probably the most effective cycle among any 2 warmth reservoirs. So, even though you've a truly gasoline, the potency might pass down however it's going to nonetheless be one hundred% of the Carnot potency. However, I query whether or not a truly gasoline may also be real reversible and keep the features of a Carnot cycle.

Combustion engine is not quite equivalent to Carnot engine.

The formula (T-To)/T does not apply here.

For example if you consider AA battery, operating

at temperature, equal to ambient temperature, then

its efficiency must be what? zero?