Heat of Combustion Question (1 Viewer)

boxhunter91

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LOL Aerath you raped me.
+ Rep.
Would we still get 2/3?
EDIT: You must spread reputation yadadada
 

Gibbatron

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I don't think so. It means you have to find out how many moles of ethanol have been combusted using:

1367=mCAT/n

you can find mCAT so you solve for the moles of ethanol and then multiply this value by 2 because if half the heat is lost then you need twice as much to give the same total heat. Using this you can find the mass, etc, etc
 

gurmies

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I started racking my brain over this bullshit, do I divide or multiply by 2...but then I realised, fuck it, I'm multiplying final mass by 2.
 

captainvagina

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this is the solution to the combustion question:

Heat = -0.21 x 4.18 x 10^3 x 65
=57057J
=57.057kJ

n=57.057/1367
=0.0417mol
then you times this number by 2 as twice as many moles of ethanol is needed to achieve the heat energy of 57.057
=0.0834mol

then using n=m/Mm

m= 0.0834mol x ((12 x 2) + (1 x 6) + 16)
= 3.84g
 

flyace

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yeh i doubled the value since the mass of ethanol burtn has to make up for the heat lost in order to still raise the temp. of the water to 65 degrees.

The question dint say nethin about the delta T changing altho heat was lost

i got 3.84 g
 

humdiddle

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Heat 210g of water by 65ºC using ethanol

specific heat capacity= 4.18

heat of combustion of ethanol = 1367KJ/mol

∆H=mc∆t

= 210 x 65 x 4.18

= 57057 J (57.057 KJ)

Since half heat is lost to surroundings, only 683.5 KJ/mol of heat was utilized from the ethanol.

so 57.057/moles of ethanol = 683.5

...since n=mass/molecular mass

57.057/(mass/molecular mass) = 683.5

therefore mass/molecular mass = 57.057/683.5

= 0.083477688....

so mass of ethanol = 0.083477688 x molecular mass (which for ethanol is 46)

therefore, mass of ethanol = 0.083477688 x 46

= 3.8 g (2 s.f) --> which is the corrent amount of significant figures based on the question.
 
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hermand

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so could you just divide the initial energy output by two instead of multiplying by two in the end? because only half of the energy of the ethanol is getting to the water?

pretty sure i got around 3.8. don't remember but yeah.
 

Aerath

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so could you just divide the initial energy output by two instead of multiplying by two in the end? because only half of the energy of the ethanol is getting to the water?

pretty sure i got around 3.8. don't remember but yeah.
Yeah, that's correct. :)
 

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