Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 16 - Temperature and Heat - Problems and Conceptual Exercises - Page 570: 85

Answer

(a) $0.21KW$ (b) $0.42KW$

Work Step by Step

(a) We know that $\frac{Q}{t}=\frac{KA\Delta T}{L}$ We plug in the known values to obtain: $\frac{Q}{t}=\frac{(0.60W/m-K)(1.40m^2)(3K)}{1.20\times 10^{-2}m}$ $\implies \frac{Q}{t}=210W=0.21KW$ (b) We know that the rate of transfer of the heat is doubled, that is, $420W=0.42KW$ when the temperature difference is doubled.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.