Molar Heat of Combustion of Ethanol (1 Viewer)

[bored of sc]

1. The molar heat of combustion of ethanol is 1360kJ/mol. What will be the final temperature if 250g of water at 20^oC is heated by the combustion of 2.0g of ethanol if there is a 30% heat loss to the surroundings.

I got all sorts of answers. The most realistic answer I got was 25.4^oC which I think is wrong. I need some help with the step-by-step process.

2. Arrange the following fuels in order of increasing amount of carbon dioxide per kilojoule of energy released. For each substance the amount of heat energy released in kJ/mol during combustion is given in brackets:
- coal, CH (455)
- methanol, CH₃OH (726)
- octane (petrol), C₈H₁₈ (5460)
- ethanol, C₂H₅OH (2021)
- methane, CH₄ (890)

This question seems fairly straight forward but I don't know where to start. Are equations required here?


Please help with these two questions. Cheers.

 

[gurmies]

The first question i'm not too sure what exactly to do with the 30% heat loss to the surroundings =(

For the second one, I think if you balance the equations you can check the mole ratios??

2CH + 3O2 ----------> 2CO2 + 2H2O [1]

2CH3OH + 3O2 ----------> 2CO2 + 4H2O [2]

2C8H18 + 25O2 -----------> 16CO2 + 18H2O [3]

C2H5OH + 3O2 ------------> 2CO2 + 3H2O [4]

CH4 + 2O2 -------------> CO2 + 2H2O [5]

Now if i'm correct, you can start comparing like so:

[1] For every 1 kJ of CH combusted, 1/455 Moles of CO2 is released

[2] For every 1kJ of CH3OH combusted, 1/726 Moles of CO2 is released

[3] For every 1kJ of C8H18 combusted, 8/5460 Moles of CO2 is released

[4] For every 1kJ of C2H5OH combusted, 2/2021 Moles of CO2 is released

[5] For every 1kJ of CH4 combusted, 1/890 Moles of CO2 is released

So in increasing amount of CO2 released:

Ethanol, Methane, Methanol, Octane, Coal

I might be extremely incorrect but that seems logical to me?

 

[jackc91]

1. The molar heat of combustion of ethanol is 1360kJ/mol. What will be the final temperature if 250g of water at 20^oC is heated by the combustion of 2.0g of ethanol if there is a 30% heat loss to the surroundings.

I'm not so sure about this but it might still help.

Moles of ethanol = 2/46.07
= 0.043412... mol

Energy released = 0.043412... * 1360
= 59.04059... kJ

Energy absorbed by water = 59.04.... * 0.7 (30% loss)
= 41.3284.... kJ

Therefore using deltaH = mCdeltaT

41.3284.... = 0.25 *4.18 * deltaT
deltaT = 39.5487
= 39.5 degrees

Therefore final temp = 59.5 degrees

 

[bored of sc]

I don't have the answers unfortunately, but you guys seem to be correct. So thanks. What you guys have posted IS logical, I can fully understand where you are coming from.

 

[jackc91]

Please post when and if you get answers. Thanks.

 

[Fortify]

EDIT: Nvm, mis-read.