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 90: 100


The final temperature is $80.78^{\circ}C$

Work Step by Step

Use the equation $q=mcΔT$, where q represents the heat that flows into a system to increase its temperature, m represents the mass of the substance, 'c' represents the heat capacity of the substance and ΔT represents the change in temperature. We can rearrange and solve for the change in temperature, (i.e. ΔT) First we must convert the heat unit from kWh to J by multiplying by $3.6\times10^{6} J $ because $1 kWh=3.6\times10^{6} J $ so $9.4\times10^{-2} kWh \times 3.6\times10^{6} J= 338400 J.$ Also the heat capacity of water is 4.184. Lastly since we are told we have 1.45 L of water or 1450 mL, we know the mass to be 1450 grams since the density of water is 1 g/mL. $q=mcΔT$ $ΔT=\frac{q}{mc}$ $ΔT=\frac{338400 J}{1450 g\times4.184} = 55.78^{\circ}C$ Therefore, the temperature change is $55.78^{∘}C$ and since $ΔT=Tfinal−Tinitial$, we can solve for the final temperature since we know the initial temperature to be $25^{∘}C$ as its given in the question. Thus, $Tfinal=ΔT+Tinitial$ so $55.78^{\circ}C + 25^{∘}C$$=80.78^{\circ}C$
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