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

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

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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.

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