Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.
ISBN 10: 1133610668
ISBN 13: 978-1-13361-066-3

Chapter 1 - The Foundations of Chemistry - Exercises - Heat Transfer and Temperature Measurement - Page 39: 64

Answer

a) $9.12\cdot10^5\ J$ b) $11.8°C$

Work Step by Step

a) Use the sensible heat equation for the rocks to calculate the heat: $q=mc\Delta T$ $q=(69.7\ kg\cdot\frac{1\ kg}{10^3\ g})\cdot 0.818\ J/g.°C\cdot (41.0°C-25.0°C)$ $q=9.12\cdot10^5\ J$ b) First, calculate the mass of air in g: $d=\frac mV$ $m=(2.83\cdot10^5\ L\cdot \frac{10^3\ mL}{1\ L})\cdot 1.20\cdot10^{-3}\ g/mL=3.40\cdot10^5\ g$ Calculate the heat released by the rocks while cooling from 41.0°C to 30°C: $q=mc\Delta T$ $q_r=(69.7\ kg\cdot\frac{1\ kg}{10^3\ g})\cdot 0.818\ J/g.°C\cdot (30.0°C-41.0°C)$ $q_r=-6.27\cdot10^5\ J$ Neglecting heat losses, all the heat lost by the rocks is absorbed by the air: $q_r+q_a=0$ $q_a=+6.27\cdot10^5\ J$ Calculate the final temperature of the air from the sensible heat equation: $q=mc\Delta T$ $6.27\cdot10^5\ J=3.40\cdot10^5\ g\cdot 1.004\ J/g.°C\cdot (T_f-10°C)$ $1.84=T_f-10$ $T_f=11.8°C$
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