Calculate the mass of ethanol that must be burnt to increase the temperature of 150g of water by 75 degrees celsius, if exactly 75% of the heat released by this combustion is lost to the surroundings. Also it is given that the heat of combustion of ethanol 1367 kJ.mol^-1 and the specific heat capacity of water is 4.18x10^3 J.kg^-1.K^-1.
what are two ways to limit heat loss from this experiment ?
How much heat do you need?
q = mass H2O x specific heat H2O x dT
q = 150g x 4.18 J/g x 75 = approx 47000 J but you need to go through all of these calculations and redo them for better accuracy.
You will lose 75% (or keep 25%) so we must increase that to approx 47,000/0.25 = approx 188,000. (Check that out to make sure it's right. If we have 188,000 and lose 75% we lose 188,000 x 0.75 = 141,000. That leaves us with 188,000-141,000 = 47,000 J that we can use.)
C2H5OH + 3O2 --> 2CO2 + 3H2O + 1367 kJ/mol.
1367 kJ/mol x #mol = 188 kJ
Solve for #mol and convert mols to grams ethanol. g = mols x molar mass.
posted by DrBob222