Question
Asked Nov 13, 2019
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90.0g of Ethyl alcohol is mixed with 200g of water.  If the mixture is heated from 20 degrees celcius from 78 degrees celcius, what is the energy required to heat the mixture?  ( C ethanol = 2.3 kj/kg C) (CH2O = 4.18 kj/kg C)

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Expert Answer

Step 1

Given,

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Mass of Ethyl Alcohol,m 90 gm 0.09 Kg Mass of Water, m, =200 gm 0.20 Kg Specific heat of ethanolC 2.3 kJ/Kg°C Specific heat of water,C, 4.18 kJ/Kg°C Initial Temperature,7 20° C Final Temperature,7 =78° C

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Step 2

Energy required to heat is the mixture is equal to sum of heat required by each component of mixture,

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о-а +а о-тсдт +тс,Ат

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Step 3

By substituting the given values the ...

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Q m,GATm,C,AT -(0.09)(2.3x10')(78-20)+(0.20)(4.18x10')(78-20) Q 12006 48488 Q=60494 J Q60.49 kJ

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