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 7

Q

A

$250 \textrm{ kg/hr}$

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VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. We're gonna take our usual formula for heat transfer, *Q* equals *mc* times change in temperature, and divide both sides by time. And we can solve for kilograms per hour or mass per time by dividing both sides of this by *C Δ* temperature. And that gives us mass per time is the heat per time divided by specific heat capacity times change in temperature. So, that's 32,000 kilojoules per hour or times 10 to the 3 joules per hour. And then divided by 4,186 joules per kilogram Celsius degree and then times by the change of temperature, final temperature of 42 degrees Celsius minus initial temperature of 12 degrees Celsius. And that gives about 250. And the units are kilograms per hour. The joules cancelled here the Celsius degrees cancelled here. And having 1 over kilograms in the nominator is the same as having kilograms in the numerator. And you can multiply top and bottom by kilograms to see how that works here. So, this is kilograms per hour or mass per time of water heated up.

COMMENTS

By Iamurfather on Fri, 3/2/2018 - 12:50 AM

If the question asked for amount of water, is it okay to end with kg of water? Thank You.

By Mr. Dychko on Sun, 3/11/2018 - 11:19 PM

Hi chaegyunkang, it looks to me like the question is asking for mass per time, which means there needs to be units of mass divided by time, which could be kg / hr, but not just kg.

All the best,

Mr. Dychko

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