College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 11 - Heat - Learning Path Questions and Exercises - Exercises - Page 412: 11

Answer

$5802KJ$

Work Step by Step

We can determine the total heat absorbed by the engine as follows: As $Q_{Al}= m_{Al}C_{Al}\Delta T_{Al}$ We plug in the known values to obtain: $Q_{Al}=(25\times 10^3) (0.900) (120-20)$ $\implies Q_{Al}=2250KJ$ Similarly $Q_{Fe}=m_{Fe}C_{Fe}\Delta T_{Fe}$ $\implies Q_{Fe} =(80\times 10^3) (0.444) (120-20)=3552KJ $ Now $Q=Q_{Al}+Q_{Fe} $ We plug in the known values to obtain: $Q=2250+3552=5802KJ$
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