Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 14 - Heat - Problems - Page 409: 20

Answer

The specific heat of the substance is $341~J/kg~C^{\circ}$

Work Step by Step

The energy $Q$ required to raise the temperature of a substance is: $Q = mc~\Delta T$ We can find the heat energy gained by the aluminum, water, and glass. $Q = m_a~c_a~\Delta T+ m_w~c_w~\Delta T+m_w~c_w~\Delta T$ $Q = \Delta T~(m_a~c_a+ m_w~c_w+m_w~c_w)$ $Q = (35.0^{\circ}C-10.5^{\circ}C)~[(0.105~kg)(900~J/kg~C^{\circ})+(0.185~kg)(4186~J/kg~C^{\circ})+(0.017~kg)(840~J/kg~C^{\circ})]$ $Q = 21,638~J$ The heat energy lost by the substance will be equal in magnitude to the heat energy gained by the aluminum, water, and the glass. We can find the specific heat $c_s$ of the substance. $Q = m_s~c_s~\Delta T$ $c_s = \frac{Q}{m_s~\Delta T}$ $c_s = \frac{21,638~J}{(0.215~kg)(330^{\circ}C-35.0^{\circ}C)}$ $c_s = 341~J/kg~C^{\circ}$ The specific heat of the substance is thus $341~J/kg~C^{\circ}$.
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.