Energy problem - ice melted by burning ethanol

AI Thread Summary
The discussion revolves around calculating how much ice at 0°C can be melted by the heat produced from burning 10.0g of ethanol. The combustion reaction of ethanol is provided, and the heat of formation is calculated to be -1235.6 kJ, indicating the energy released during combustion. The user initially struggles with converting this energy into grams of melted ice but eventually realizes the need for the latent heat of fusion of ice and the energy released per gram of ethanol. The correct approach involves equating the heat needed to melt the ice with the heat released from burning ethanol. The user confirms their calculations after correcting the energy value used.
philtered
Messages
8
Reaction score
0

Homework Statement


How many grams of ice at 0°C could be melted by the heat produced when 10.0g of ethanol is burned?

Homework Equations


So I wrote out the combustion reaction:
C2H5OH +3O2 ---> 2CO2 + 3H2O

The Attempt at a Solution


First I though I had to find heat of formation for that reaction, which I found to be -1235.6kJ (So this would be the heat given off when ethanol is burned correct?)
Hf=products-reactants
Hf=[-393.6(2)+-242(3)]-[-277.6+0]
Hf=-1235.6kJ

I wasn't sure if I had to use that number or if I had to just pull the Hf number off my chart for ethanol (-277.6kJ)

I have the answer given to me, but I'm stuck at converting my kJ to g of water. (Answer is 802g)

Tried: 1235.6kJ[1mol/6.03kJ][18.02g/1mol]
 
Physics news on Phys.org
You need to constants. The latent heat of fusion of ice (let us denote it by L_{f}) and the energy released by burning a unit mass of ethanol (let us denote it by q). Then, you simply have:

Heat needed to melt mass m(\texrm{ice}) = L_{f} \cdot m(\textrm{ice})

Heat released by burning mass m(\textrm{ethanol}) = q \cdot m(\textrm{ethanol})

These are equal, so, you have an equation for m(\textrm{ice}):


<br /> m(\textrm{ice}) = \frac{q}{L_{f}} m(\textrm{ethanol})<br />
 
Thanks for the help. It worked out with that. Also the q value I was using was wrong, had forgot to take into account the 10g of ethanol being burned.
 
cool. am i right to assume you're a chem student?
 
Yeah I am. Right now just the high school course, next year I'm doing Physics with a minor in Chem
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
Back
Top