Heat of Combustion of Ethanols (1 Viewer)

[Zatts]

Hey I was just wondering if anyone could verify that what I'm doing is right, cause I'm kinda unsure that I am doing the right thing.

I have a spirit burner full of ethanol, and a can with 200mL of water.

Weight of spirit burner before: 241.52g
Weight of spirit burner after: 240.90g
Therefore, change in mass is 0.62g

Temp of water before: 17°C
Temp of water after: 29°C
Therefore, change in temp is 12°C

So, using the equation ∆H = mC∆T:

∆H = (0.2) x (4.18x10^3) x (12)
∆H = 10,032 J

Number of moles = 0.62g / 46.086 (Molar Mass)
Number of moles = 0.0134

Heat per mole = 10,032 / 0.0134
Heat per mole = 7486.567 kJ/mole

Please, tell me if I did anything wrong there. I don't have the answer to this question, and I really need to make sure I'm doing these types of questions right.

Any help would be greatly appreciated!

 

[rnitya_25]

firstly, i thought it was - mC(delta)T....exothermic

 

[Dreamerish*~]

∆H = (0.2) x (4.18x10^3) x (12)
∆H = 10,032 J

I see you have converted the 200 mL to 0.2 L. When you do that, the energy comes out as kJ. If you want your energy in joules, leave your mass in grams. I don't know why your heat of capacity of water is 4.18 x 10³, but I've always just had 4.18.

Rememeber that g goes with J, and kg goes with kJ.

Your equation would read, according to my methods of calculation:

∆H = 200 x 4.18 x 12
∆H = 10,032 J

We have the same answer because you had the 200 mL of water converted to litres, but timed 4.18 by 1000. Could you tell me why your heat of capacity of water is 4.18 x 10³? I've never seen it done that way before. But your answer is correct.

firstly, i thought it was - mC(delta)T....exothermic

Yes, the equation is ∆H = mC∆T.

Don't worry about the negative sign, because you know that the reaction is exothermic. ∆H is either the amount of energy released or absorbed - depending on the reaction. It's true that when ∆H is negative, then the reaction is exothermic, but for calculation purposes regarding a reaction that is known to be either endo or exo, it doesn't matter.

 

[Zatts]

Could you tell me why your heat of capacity of water is 4.18 x 10³? I've never seen it done that way before.

Well we were given that value as the specific heat capacity of water. And also, looking at the back of the periodic table we got in the half-yearlies, it says 4.18 x 10^3 J kg^-1 K^-1.

But yeah thanks for your help !

 

[taxman]

Yeah, I've noticed that in the sheet we're given for exams, specific heat capacity is noted as 4.18x10^3....I don't think it makes too much of a difference marking wise...but I have always written down 4.18.

I've always done the equation including heat absorbed by the copper container...so it looks something like

(delta)H = mass x (heat absorbed by container + heat absorbed by water)
= 0.62g x ( 0.38 x 12 + 4.18 x 12)

I've seen it done many times using both ways. I assume there's no correct or incorrect way of doing things. In my experience, using heat gained by container usually brings in closer to SI values than without it.

 

[samuelblayden]

hey I just got out of yr11 chem a few secs ago and we weere talking about this

 

[rama_v]

I always do all of these questions the same way:

First you find how many grams of substance (i.e. ethanol or another alkanol) was used. Convert it to moles. Then plug in the numbers in to the formula, i always use 4.2 J/g/K (instead of 4.2 x 10³) because it means i dont have to convert the grams of water to kilograms.
So therefore you can then write x mol of this alkanol stores y J
therefore 1 mol of this alkanol stores ? J

so x/y = 1/? you know the values for x and y, so sub in to find for the ?

then divide by 1000 to get kJ/mol.

 

[-Swifty-]

Yeah, I've noticed that in the sheet we're given for exams, specific heat capacity is noted as 4.18x10^3....I don't think it makes too much of a difference marking wise...but I have always written down 4.18.

I've always done the equation including heat absorbed by the copper container...so it looks something like

(delta)H = mass x (heat absorbed by container + heat absorbed by water)
= 0.62g x ( 0.38 x 12 + 4.18 x 12)

I've seen it done many times using both ways. I assume there's no correct or incorrect way of doing things. In my experience, using heat gained by container usually brings in closer to SI values than without it.

yeah just make sure u look at the units on the data sheet..... i cant remember off by memory the units for it, but yeah just check them which someone mentioned b4.

thou in my trials i put the mass as the change in ethanol.... doh! stupid of me

 

[Bokky]

i still dont get this crap, whats the whole aim of this experiment?? i have 4 text books on chemistry, i looked at hsc online, and i looked here, none have the bloody steps of what to do after u calculate the stupid H=mCT or watever. any1 out there with an answer ????????

 

[Dreamerish*~]

i still dont get this crap, whats the whole aim of this experiment?? i have 4 text books on chemistry, i looked at hsc online, and i looked here, none have the bloody steps of what to do after u calculate the stupid H=mCT or watever. any1 out there with an answer ????????

The aim is to calculate the different heat of combustion of alkanols.

The formula is ΔH = MCΔT, not "H=mCT or whatever".

ΔH = the amount of energy released
M = mass of water used (1 g would equal 1 mL)
C = molar heat capacity of water, which is usually given as 4.18 or 4.2
ΔT = the change in temperature of water

After you work out ΔH, you convert it to Joules per gram or per mole.

If you have a calculations question you need help with, post it up.

 

[Bokky]

"After you work out ΔH, you convert it to Joules per gram or per mole. "

And how do u do that?

for example you have water which is say 50ml, the temp change is 10C and u put it into the equation:

H = 50 x 4.18 x 10
H = 2090 Joules per gram?
now wat do u do?

 

[Dreamerish*~]

"After you work out ΔH, you convert it to Joules per gram or per mole. "

And how do u do that?

for example you have water which is say 50ml, the temp change is 10C and u put it into the equation:

H = 50 x 4.18 x 10
H = 2090 Joules per gram?
now wat do u do?

What did you burn? Ethanol? How much did you burn?

Okay, let's assume you burnt 5 g of ethanol (just assuming to give you an example - don't use these figures). Work out the molar mass of ethanol (C₂H₅OH) - which is 46 g.

From your experiment you got 2090 J from burning 5 g.

How many moles of ethanol is that?

5/46 = 0.108695652 mol

Which means you got 2090 J from burning 0.108695652 mol of ethanol.

Converting it to J per mole, 2090/0.108695652 = 19228 J

Therefore the molar heat of ethanol, in your case would be 19228 J mol^-1

Does that help?

 

[Bokky]

yes thankk u very much

 

[~GrOoVy~]

so in the exam do we take 'c' as 4.18 X 10^3 or just 4.18???

 

[Dreamerish*~]

so in the exam do we take 'c' as 4.18 X 10^3 or just 4.18???

Ok, assuming you want your ∆H to be in joules, which is usually is:

Well we were given that value as the specific heat capacity of water. And also, looking at the back of the periodic table we got in the half-yearlies, it says 4.18 x 10³ J kg^-1 K^-1.

Which means you give your mass of water in kg. That is, 200 mL of water would be 0.2 kg.

However, if you convert the mass of water to g (200 mL becomes 200 g) and that is the unit you put into ∆H = MC∆T, then you use 4.18.

∆H = (0.2) x (4.18x10^3) x (12)
∆H = 10,032 J

∆H = 200 x 4.18 x 12
∆H = 10,032 J

As you can see, Zatts and I both ended up with the same answer.