College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 14 - Problems - Page 535: 37


The required heat is $3.07\times 10^6~J$

Work Step by Step

We can find the heat required: $Q = m~c_i~\Delta T+m~L+m~c_w~\Delta T+m~V+m~c_s~\Delta T$ $Q = m~(c_i~\Delta T+L+c_w~\Delta T+V+c_s~\Delta T)$ $Q = (1.0~kg)[(2100~J/kg~C^{\circ})(20.0~C^{\circ})+(333\times 10^3~J/kg)+(4186~J/kg~C^{\circ})(100.0~C^{\circ})+(2260\times 10^3~J/kg)+(2010~J/kg~C^{\circ})(10.0~C^{\circ})]$ $Q = 3.07\times 10^6~J$ The required heat is $3.07\times 10^6~J$
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.