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: 58

Answer

28 minutes

Work Step by Step

We first must find how much energy is needed to bring the water to a boil: $Q = mc\Delta t = (2.5)(4184)(90)=941,400\ J \approx 941 \ kJ$ Recall, the energy supplied is equal to the integral of P with respect to time. Note, we must substitute a "dummy variable" in for t, for the same variable cannot be in both the expression and in the bounds of the integral. Thus, we find: $ 941 = \int_0^t 1.1 + 2.3 s$ $ 0 = 1.15t^2 + 1.1t -941 $ Using either a calculator or the quadratic formula, we obtain: $ t = 28 \ min$
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