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

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Homework Help Overview

The discussion revolves around calculating the heat absorption in an automobile cooling system containing 18 liters of water, specifically examining the temperature change from 20°C to 90°C.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the heat absorbed using the formula Q=mcΔT and presents their calculations. Some participants question whether all necessary information has been provided and express uncertainty about the book's answer.

Discussion Status

The discussion is ongoing, with participants exploring the calculations and questioning the accuracy of the textbook answer. There is no explicit consensus on the correctness of the original poster's calculations or the textbook's response.

Contextual Notes

Participants note that the calculations are based solely on the information provided, and there is a suggestion that the textbook may contain an error.

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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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