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: 34

Answer

The mass of the ice cube was 9.77 grams.

Work Step by Step

We can find the heat energy lost by the water and the aluminum that was originally at room temperature. $Q = m_w~c_w~\Delta T + m_a~c_a~\Delta T$ $Q = (0.310~kg)(4186~J/kg~C^{\circ})(20.0~^{\circ}C-17.0~^{\circ}C) + (0.085~kg)(900~J/kg~C^{\circ})(20.0~^{\circ}C-17.0~^{\circ}C) $ $Q = 4122.5~J$ This is the heat energy required to raise the temperature of the ice, melt the ice, and raise the temperature up to $17.0^{\circ}C$. We can find the mass of the ice. $Q = m~c_i~\Delta T_i+m~L + m~c_m~\Delta T_w$ $m = \frac{Q}{c_i~\Delta T_i+L + c_m~\Delta T_w}$ $m = \frac{4122.5~J}{(2100~J/kg~C^{\circ})(8.5~C^{\circ})+3.33\times 10^5~J/kg + (4186~J/kg~C^{\circ})(17.0~C^{\circ})}$ $m = 0.00977~kg = 9.77~g$ The mass of the ice cube was 9.77 grams.
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