Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 17 - Exercises and Problems: 49

Answer

Please see the work below.

Work Step by Step

We know that $Q_{water}=Q_{boiling}$ $\implies mc\Delta T=(\frac{1}{10})mL_v$ $mc(100-T_i)=(\frac{1}{10}mL_v)$ $c(100-T_i)=(\frac{1}{10}L_v)$ We plug in the known values to obtain: $(4184)(100-T_i)=(\frac{1}{10})(2257000)$ This simplifies to: $T_i=46^{\circ}C$
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