Giancoli's Physics: Principles with Applications, 7th Edition

14

Heat

Change chapter14-1: Heat as Energy Transfer

14-3 and 14-4: Specific Heat; Calorimetry

14-5: Latent Heat

14-6 to 14-8: Conduction, Convection, Radiation

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.

Problem 10

Q

A

$1.7 \times 10^3 \textrm{ J/kg C}^\circ$

In order to watch this solution you need to have a subscription.

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. We can solve this formula for heat gain by a substance for *c* the specific heat capacity by dividing both sides by mass times change in temperature. So, *c* is *Q* over *m ΔT*, and it absorbs 135 times 10 to the 3 joules divided by 4.1 kilograms times 37.2 degrees Celsius, final temperature, minus 18 degrees Celsius, initial temperature. And we get about 1.7 times 10 to the 3 joules per kilogram Celsius degree is the specific heat capacity of this metal.

Giancoli Answers, including solutions and videos, is copyright © 2009-2024 Shaun Dychko, Vancouver, BC, Canada. Giancoli Answers is not affiliated with the textbook publisher. Book covers, titles, and author names appear for reference purposes only and are the property of their respective owners. Giancoli Answers is your best source for the 7th and 6th edition Giancoli physics solutions.