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: 27

Answer

The mass of the steam required is 0.180 kg

Work Step by Step

We can find the heat energy required to melt the ice and raise the temperature up to $30^{\circ}C$. $Q = m~L + m~c_w~\Delta T$ $Q = (1.00~kg)(3.33\times 10^5~J/kg) + (1.00~kg)(4186~J/kg~C^{\circ})(30~C^{\circ})$ $Q = 458,580~J$ The heat energy required by the ice will be equal to the heat energy lost by the steam. $Q = m~L+m~c_w~\Delta T$ $m = \frac{Q}{L+c_w~\Delta T}$ $m = \frac{458,580~J}{(2.26\times 10^6~J/kg) + (4186~J/kg~C^{\circ})(100~C^{\circ}-30~C^{\circ})}$ $m = 0.180~kg$ The mass of the steam required is 0.180 kg.
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