Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 16 - Exercises and Problems: 73


a) 418.76 b) .09 percent higher

Work Step by Step

We know that the energy needed is given by: $Q = \int_0^{100} mc(T) dt$ Since the mass is one, this simplifies to: $Q = \int_0^{100} c(T) dt$ Plugging c(T) into this equation, we find: $Q=418,763.33 \ J \approx 418.76 \ kJ$ b) If we were to use the approximation, the value would be: $Q = mc\Delta T = (1)(4184)(100)=418,400 \ J$ The actual value is $\fbox{.0867 percent higher}$
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