Efficiency of Carnot engine

The ratio between input and out is called efficiency. The work done on machine is input and the work done by machine is output. The input of Carnot is equal to the amount of heat Q1 supplied during isothermal process.
The output is equal to the difference of work done on the system and the work done by the system. During Carnot cycling engine work done in two steps. In final two steps to bring the engine in its initial stage the work is done on the system. At returning the engine in its initial stage its internal energy will be equal to zero.
ΔU = 0 therefore it observe Q1 heat in isothermal expansion process and release Q2 heat in isothermal compression and the resultant of absorb heat will be equal to the
ΔQ = Q1 – Q2
According to the 1st law of thermodynamic
ΔQ = ΔW + ΔU
Q1 – Q2 = ΔW (ΔU = 0)
Q1 = input
Q1 – Q2 = output
Efficiency = output/ input
E = ΔW/ Q1 > E = Q1 – Q2/ Q1
E = (Q1/Q1 – Q2/Q1)
E = (1 - Q2/Q1) *100 % or Q2/Q1= T2/T1
E = (1 - T2/T1) *100 %




Related problem
Efficiency of a heat engine is 20 % and does 20 joule work in each cycle find in each cycle
a. The amount of heat absorbs?
b. The amount of heat release?
Solution
Data
Work done in each cycle = output of each cycle ΔW = 200 J
Efficiency = 20 %
Amount of heat absorbs = Q1
Amount of heat release = Q2
Efficiency (E) = output/ input * 100
E = ΔW /Q1 * 100
20 = 200 / Q1 *100
Q1 = 200 * 100 /20
Q1= 1000 J heat absorbs in each cycle

Q1 – Q2 = ΔW
1000- Q2 = 200
Q2 = 1000-200
Q2 = 800 J