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: 103

Answer

The mass of the fuel required is $21.792 grams$

Work Step by Step

Using the equation $q=mcΔT$, we can solve for the heat. The specific heat capacity of water is $4.184\frac{J}{g\times^{\circ}C}$, and the mass of the water in this case is 2.5 kg or 2500 grams. Also the change in temperature of the water can be calculated by taking the final temperature and subtracting the initial temperature; $100^{\circ}C - 25^{\circ}C = 75^{\circ}C$ $q=(2500 grams×4.184\frac{J}{g\times^{\circ}C}×75^{\circ}C)=784,500 J$ Now we must convert the units of joules into kilojoules by simply dividing by 1000; $784,500 J\div 1000 = 784.5 kJ$ Now to find the mass of the fuel we can divide the heat required for the water to increase in temperature by the heat of the fuel/unit mass; $784.5 kJ\times\frac{1 g}{36 kJ} = 21.792 grams$
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