The complete combustion of 1.00 ml of octane increases the temperature by 22.7 degrees celsius. The density of octane is 0.7g/mL. What is the efficiency of the octane in heating the water?
8 answers
You have to know how much water.
Sorry. There are 250.00g of water.
find the heat of commbusion, then figure the heat from the ocatane.
Now, the water: heat to heat water:masswater*c*22
efficiency=heatforwater/heatfromoctane * 100 if you want it in percent.
Now, the water: heat to heat water:masswater*c*22
efficiency=heatforwater/heatfromoctane * 100 if you want it in percent.
I have a feeling this is just part of th question and that it refers to another question.
Increases the temperature of what 22.7 celsius? Water?
Calculate dH combustion octane.
2C8H18 + 25O2 ==> 16CO2 + 18H2O
dHrcomb(rxn) = (n*dHf products) = (n*dHf reactants) = ? = heat of combustion for octane. This is the amount of heat available (theoretical I've called it below).
Use density to convert 1.00 mL octane to grams. Mass of water octane heated is not given.
q = mass H2O x specific heat H2O x delta T = this is the actual heating accomplished.
Effeciency = [(actual heating)/theor heating)]*100 = ?
Increases the temperature of what 22.7 celsius? Water?
Calculate dH combustion octane.
2C8H18 + 25O2 ==> 16CO2 + 18H2O
dHrcomb(rxn) = (n*dHf products) = (n*dHf reactants) = ? = heat of combustion for octane. This is the amount of heat available (theoretical I've called it below).
Use density to convert 1.00 mL octane to grams. Mass of water octane heated is not given.
q = mass H2O x specific heat H2O x delta T = this is the actual heating accomplished.
Effeciency = [(actual heating)/theor heating)]*100 = ?
But @DrBob222 where would I find delta T? Also, mass of water is 250 grams.
delta T is 22.7. Right?
I'm getting really high negative percentage numbers.
Bob Pursley's answer and my are are the same. Post your work if you're still having trouble.
1 answer