Heat of Combustion Question (1 Viewer)

sam5

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Tell me if i did this wrong.

Heat = -0.21 x 4.18 x 10^3 x 65

= 57.057 KJ

I then divided this value by 2, because it said half the heat was lost to the environment.

Therefore, Heat = 28.5285

As HoC = Heat on moles

Therefore ; 1367 (given) = 28.5285 divided by m/(12.01 x 2 + 1.008 x 6 + 16)

Mass of Ethanol = 0.96 g (2.d.p.)
 

boxhunter91

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I had the other answer then changed to this.
One of these answers are correct..
Either way you will get at least 2/3 if not 3.
 

kanux

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don't you multiply by two, since the heat is lost to surroundings? :S
 

sam5

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I had the other answer then changed to this.
One of these answers are correct..
Either way you will get at least 2/3 if not 3.
thats good mate. :)

I need some good news to break up my legal studies study :(
 

sam5

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don't you multiply by two, since the heat is lost to surroundings? :S
why would i multiply the heat value by 2, when the heat that is absorbed by the water is only half of that theoretical value.

Im not doubting u, i just want an explanation if ur correct.
 

Blaz-357

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i just calculated it normally, as u would, then multiply my final mass by 2, since i would need twice the mass to make up for the half lost to the surrounding.
 

kanux

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i forgot how the question went
but aren't we trying to find out how many grams of ethanol needed to get 57.057 kJ?


since we know half the heat is lost to surroundings, we double that value. so we're finding 57.057x2 kJ since half is lost to surroundings.
 

vpa2891

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I multiplied q value by 2 cuz the q u calculated was the heat absorbed by water. There was the same amount of heat lost to surrounding so the total heat was 2 times that...
 

imalda

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don't you multiply by two, since the heat is lost to surroundings? :S
i multiplied it by two as well since it said exactly half was lost to surroundings. don't remember what i got exactly though. something between 1 and 3 LOL my memory is like shit XD
 

zeleboy

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why would i multiply the heat value by 2, when the heat that is absorbed by the water is only half of that theoretical value.

Im not doubting u, i just want an explanation if ur correct.
If only half is absorbed then it would make sense to double your ethanol input to make sure you get required output?
 

kanux

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i multiplied it by two as well since it said exactly half was lost to surroundings. don't remember what i got exactly though. something between 1 and 3 LOL my memory is like shit XD
lol cool, me too bout the memory...i feel old D:
anyways i stuffed that question anyways, i think i forgot to change to kJ. cos i got a ridiculously large number :/
 

sam5

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yeah i totally understand yas.

Ive done it wrong. I just thought that because only half the heat was reaching the water, i thought that i would halve the theoretical heat to calculate how much was actually heating the water.

Anyways. Thanks guys. :(
 

Gibbatron

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I think i got 5 something grams. I cant remember. Dammit if i did and everyone else got 3.

My method was to find the heat of energy like the OP did, then used that to find the moles of ethanol used:

1367 (given) = 57.xxx/n

solved for n and multiplied by 2 to account for half the heat lost, then used n to get a value for the mass. If thats the right method then i must've made a calculation error. Damn.
 

shaon0

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This is the only question i stuffed up in the exam. I feel bad now for not studying MHC.
 

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