College Physics (4th Edition)

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

Chapter 14 - Problems - Page 534: 28


The temperature of the air would increase by $57.5~C^{\circ}$, so the new temperature of the air would be $77.5~^{\circ}C$

Work Step by Step

We can find the number of moles of nitrogen: $PV = nRT$ $n = \frac{PV}{RT}$ $n = \frac{(1.01\times 10^5~Pa)(8000~m^3)}{(8.314~J/mol~K)(293.15~K)}$ $n = 3.315\times 10^5~mol$ We can find the total energy that heats the gas: $Q = (501)(110~W)(7200~s) = 3.968\times 10^8~J$ We can find the change in temperature: $Q = c~n~\Delta T$ $\Delta T = \frac{Q}{c~n}$ $\Delta T = \frac{3.968\times 10^8~J}{(20.8~J/mol~C^{\circ})(3.315\times 10^5~mol)}$ $\Delta T = 57.5~C^{\circ}$ The new temperature of the air will be $20.0^{\circ}C+57.5~C^{\circ}$ which is $77.5~^{\circ}C$.
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