heat of combustion thingy (1 Viewer)

[fwuxed]

meh, shouldn't have left this til the last minute but i had trouble figuring out one of the conquering chem questions on heat of combustion

30. how much ethanol is required to heat 350 g of h2o through 77c, if 50% of the heat released by ethanol is lost to the surrounding environment. heat capacity of water is 4.2 and heat of combustion of ethanol is 1360kj/mol.

i have

C2H5OH + 3O2 => 2CO2 + 3H2O

i mustve tried every value and calculated everything i could work out x_X
but to no avail

conquering chem's answer of 7.7g doesn't exactly explain anything either

 

[CM_Tutor]

The heat required to change the temperature of water through 77 °C is:

Δq = mcΔT = 350 * 4.2 * 77 = 113190 J = 113.19 kJ

Since efficiency is 50 %, heat required from combustion reaction = 113.19 / 0.5 = 226.38 kJ

Now, ΔH = -Δq * n(C₂H₅OH in eq'n) / n(C₂H₅OH)

So, we have -1360 = -226.38 * 1 / n(C₂H₅OH)
So, n(C₂H₅OH) = 226.38 / 1360 = 0.16645 ... mol

So, m(C₂H₅OH) = nM = 0.16645... * 46.068 = 7.668 ... g = 7.7 g (2 sig fig)

 

[fwuxed]

from the vague examples they gave me
i tried to use the q = mc.deltaT equation
and i also incorporated the n=m/M

so 3.n(h2o)=350/(2+16)=19.4444
n(ethanol)=4.861...=m/(24+16+6)

then i ended up with ridiculous values and im lost =[

 

[CM_Tutor]

Are you saying that you don't understand my answer or ...?

 

[fwuxed]

hm, kinda
lol

wondering which eqns you used in your working

hmm
oh well never mind, I'm guessing its a specific method of calculating these types of problems

but i hate to do something without fully understanding what i'm doing