Calculate Magnitude of Work from 0.22 Lawnmower Engine in 1 Second

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A lawnmower engine with an efficiency of 0.22 rejects 9900 J of heat per second, leading to the calculation of work done in one second. The formula used is e = W/Qh, where Qh is the sum of heat rejected and work done. The calculations suggest that the work done, W, is approximately 2792.31 J/s. There is a request for verification of this calculation, as a reference from cramster.com provides a different result. The discussion emphasizes the importance of significant figures and correct unit representation in the final answer.
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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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