Calculate Heat Flows in Engines Propelling a Car

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SUMMARY

The discussion focuses on calculating heat flows in an engine propelling a 1600 kg car to a speed of 22 m/s in 9.7 seconds with an efficiency of 23%. The heat flow into the engine at high temperature is calculated using the kinetic energy formula, yielding 199,587.63 J/s. The relationship between heat flow in and out is established through the efficiency equation e = 1 - (Qout/Qin), emphasizing the conservation of energy principle in thermodynamics.

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Homework Statement


A certain engine can propel a 1600 kg car from rest to a speed of 22 m/s in 9.7 s with an efficiency of 23%.

what is the rate of heat flow into the engine at the high temperature?

what is the rate of heat flow out of the engine at the low temperature?

Homework Equations



e= 1- (Qout/ Qin)

The Attempt at a Solution



for the heat flow into the engine (high temp):
K= 0.5mv^2= (0.5)(1600)(22^2)= 387200 J

then I did 387200 J/ (0.23 x 9.7)= 199587.6289 J/s (not sure if this is correct)

I'm unsure as to how to do the second part of the question. Please help! Thanks :)
 
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Right so you have heat going in and work being done and heat going out.

So by conservation of energy what equation must relate these quantities??
 
I get it! thanks for your help :)
 

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