A heat engine is 20% efficient. If it loses 800 J to the cooling system and exhaust during each cycle, the work done by the engine is:
200 J
1000 J
800 J
20 J
η=Q₁-Q₂ /Q₁
Q₂=800J
η=0.2
Q₂=Q₁ (1- η)
Q₁=Q₂/(1- η)=800/(1-0.2) =1000 J
W= Q₁-Q₂ =1000-800=200 J
Work = Energy output(Eo).
Eff. = Eo/(Eo+800) = 0.20
Eo = 0.2(Eo+800)
Eo = 0.2Eo + 160
0.8Eo = 160
Eo = = 200 J.
To find the work done by the engine, we need to first calculate the input energy or heat supplied to the engine during each cycle.
We know that the engine is 20% efficient, which means that 20% of the input energy is converted into useful work, while the remaining 80% is lost as waste heat.
Let's assume the input energy or heat supplied to the engine during each cycle is represented by Q.
According to the problem, the engine loses 800 J to the cooling system and exhaust during each cycle. This loss of energy is considered waste heat.
Therefore, we can equate the efficiency of the engine to the ratio of the useful work done (W) to the input energy (Q), and express it as a decimal:
Efficiency = W / Q = 0.20
We need to solve for Q:
Q = W / 0.20
Now, we know that the waste heat is 800 J, so the useful work done is equal to the input energy minus the waste heat:
W = Q - 800
Substituting the expression for Q:
W = (W / 0.20) - 800
Simplifying the equation:
(W / 0.20) - W = 800
Multiplying both sides by 0.20:
W - 0.20W = 160
Combining like terms:
0.80W = 160
Dividing both sides by 0.80:
W = 160 / 0.80 = 200
Therefore, the work done by the engine is 200 J.
So, the correct answer is 200 J.