Help! - Molar heat combustion of ethanol (1 Viewer)

[ShAzABoB]

i still dont know how to work them out... my textbook only shows you one where it is assumed that there is no heat loss to the environment... most Q i come across say theres 50% of the heat lost... what do you do then? can someone please explain?

 

[onebytwo]

i suppose youve done the exp. relating to this heat of combustion stuff, if you did it should help you understand everything about it.
you are basically combusting (if thats a word) a known mass of ethanol which we can then tranfer into number of moles. and because we are allowing the heat from the combustion heat the water(which has a known specific heat capacity), we can then use the formula to work out the delta H,
does that even answer your question....?
btw what text book are u refering to?

 

[ShAzABoB]

im using conquering chemistry 4th edition... lol i dont really understand...

 

[kazan]

ok so first thing to do is rember the formula


/\H = MC/\T


with M = mass of water
C being the heat capacity of water (4.18x10^3)
/\T being the change in tempreature


so, work this equation out, then divide by the mass burnt.

then times that amount by the molar mass


(/\H / MM ) x M

MM being the molar mass
M being mass
/\H being change in enthalpy( the first equation)

 

[mitch_f1]

Okay, might as well dump this here:

In the 2001 HSC, just reading through examiners notes, and it says that people got marked down for not using the heat capacity of water (in question 17 c) as defined in the data sheet (4180 J kg^-1 K^-1). So I did the calculation using it:

H = -mc*change*temp
=-250 x 4.18x10^3 x (59-19)
=-41800000 J kg^-1 K^-1

How do I convert this to kJ g^-1 K^-1 ???

I've tried dividing it by 1000 (to get per kJ) then timsing by 1000 (to get per g), but that does nothing (obviously)....HELP

Thanks

 

[tristambrown]

post the whole q & i;ll try to work it for you

 

[pLuvia]

You would use 4.18 if the mass of the water is in grams and 4.18x10³ if it were in kg

 

[Naylyn]

The current value that the data sheet gives is C = 4.18 * 10^3 J kg^-1 K^-1

working from there;
C = 4.18 *10^3 J kg^-1 K^-1
C = 4.18 *10^3 J 1000g^-1 K^-1
C = 4.180 J g^-1 K^-1

Which is the value that you are probally mreo familiar with. If you want to use this value you have to start from the given equation and work from there, or you could change you mass to kilograms you will get the same result.

 

[Petinga]

Shazabob, by using delta H=-mct, it is the theoretical molar heat of combustion of the compound. Thus is says 50% of heat is lost then time yuour final heat of combustion by 0.5.

 

[rama_v]

Shazabob, by using delta H=-mct, it is the theoretical molar heat of combustion of the compound. Thus is says 50% of heat is lost then time yuour final heat of combustion by 0.5.

Actually if you lose 50% to the external environment then only 50% of the energy from combustion was used to heat the water. Hence the actual heat of combustion is double the one calculated.

 

[Petinga]

ur rigth rama v. i wasnt thinking straight. i am ashamed coming from the guy who come first in all sciences. i was do this before exams though

 

[bizadfar]

(/\H / MM ) x M

MM being the molar mass
M being mass
/\H being change in enthalpy( the first equation)

I think u have the brackets wrong there man.
Isn't it /\H / (MM x M)?

basically Delta H / n

 

[~shinigami~]

I think u have the brackets wrong there man.
Isn't it /\H / (MM x M)?

basically Delta H / n

Nah man, Kazan is correct. But no biggie, it's an easy mistake to make.

Like you said:

ΔH / n

But n = m / MM ,

Hence ΔH / m / MM and of course if you divide a fraction by a fraction, you "tip and times",

Thus you get (ΔH / 1) x (MM / m) and when simplified gets you:

(ΔH x MM) / m

I think that's right.

 

[amandagirgenti]

Hey, just wanted to say THANK GOD 4 U!! I was so confused b4 i read ur post on how 2 calculate the molar heat of combustion of alkanols. I knew i had 2 work out the /\H but after that i got a bit lost lol thanx 4 saving my life!!