Introductory Chemistry (5th Edition)

Published by Pearson
ISBN 10: 032191029X
ISBN 13: 978-0-32191-029-5

Chapter 3 - Matter and Energy - Exercises - Cumulative Problems - Page 91: 105


The mass of the water that must evaporate from the skin to cool it by $0.50^{\circ}C$ is $77.87 grams$

Work Step by Step

Using the equation $q=mcΔT$, we can solve for the heat. The mass of the water is unknown in this case. The mass of the water body is given and is 95 kg or 95000 grams. Also the change in temperature of the sweat is given to be $ 0.50^{\circ}C$ $q=(95000 grams×4.0\frac{J}{g\times^{\circ}C}×0.50^{\circ}C)=190000 J$ Now we must convert the units of joules into kilojoules by simply dividing by 1000; $190,000 J J\div 1000 = 190 kJ$ Now to find the mass of the water we can divide the heat required for the sweat to decrease in temperature by the heat of the water/unit mass; $190 kJ\times\frac{1 g}{2.44 kJ} = 77.87 grams$
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