Calculating Heat Absorption in an Automobile Cooling System with 18L of Water

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The discussion revolves around calculating the heat absorption of an automobile cooling system containing 18 liters of water, with a temperature increase from 20°C to 90°C. The calculations show that the mass of the water is 18 kg, and using the formula Q=mcΔT, the heat absorbed is calculated to be 5,274,360 J. However, there is a discrepancy with the book's answer of 4.7 x 10^6 J, leading to questions about the accuracy of the book. Participants in the discussion agree that the calculations appear correct and speculate that the book may contain an error. The conversation emphasizes the importance of verifying calculations and the potential for mistakes in published materials.
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An automobile cooling system holds 18 L of water. How much heat does it absorb if its temperature rises from 20oC to 90oC?

V= 18 L=0.018 m3
m=pV=(1.0*103 kg/ m3 )(0.018 m3 )= 18 kg

Q=mcT=(18 kg)(4186 J/kg*Co)( 90o-20o)= 5,274,360 J

The answer in the book says it should be 4.7*10^6, but I don't know how they derived to that answer. Any help
 
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Any help from anybody?
 
Are you sure that is all the information given? Your calculations seems right to me.
 
Yes that was all that was given...I thought I was correct also, maybe the book made a mistake...it has been known to happen. Thanks for the help!
 
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