- #1
bulbasaur88
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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.
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.