Calculate Magnitude of Work from 0.22 Lawnmower Engine in 1 Second

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The magnitude of work done by a lawnmower engine with an efficiency of 0.22, which rejects 9900 J of heat per second, is calculated to be approximately 2792.31 J/s. The efficiency formula used is e = W/Qh, where Qh is the total heat input. The correct relationship is established as Qh = Qc/(1-e), leading to the conclusion that W = eQh. The answer must be expressed in the correct units, specifically J/s, which denotes power.

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A lawnmower engine with an efficiency of 0.22 rejects 9900 J of heat every second. What is the magnitude of the work that the engine does in one second?

e = 0.22
Qc = 9900 J/s

e = W/Qh
Qh = Qc + W

e = W / Qc + W
0.22 = W / 9900 + W
0.22(9900 + W) = W
2178 + 0.22W = W
W = 2792.31 J/s?

I checked my answer on cramster.com but they have something different, but I believe my answer to be right. Can anybody verify if I am doing this correctly or not? Thank you.
 
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bulbasaur88 said:
A lawnmower engine with an efficiency of 0.22 rejects 9900 J of heat every second. What is the magnitude of the work that the engine does in one second?

e = 0.22
Qc = 9900 J/s

e = W/Qh
Qh = Qc + W

e = W / Qc + W
0.22 = W / 9900 + W
0.22(9900 + W) = W
2178 + 0.22W = W
W = 2792.31 J/s?

I checked my answer on cramster.com but they have something different, but I believe my answer to be right. Can anybody verify if I am doing this correctly or not? Thank you.
Your answer should be expressed in the correct number of significant figures. You may want to simplify your answer algebraically before plugging in the numbers:

Qh = Qc/(1-e)

W = eQh = e(Qc/(1-e))

Also, you have to put your answer for W in the correct units. J/s is a unit of power.

AM
 

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