is it just me? or did all that just majorly confuzzle other ppl too?
ive done a practice question like the orginal question posted on this thread...
the question was...
a student used each of the follwing spirit burners (being methanol, ethanol and propanol) to determine the hear of combustion of each fuel. they did this by heating 200ml of water in a beaker over each burner for 2 minutes...
the results are:
fuel methanol ethanol propanol
mass of burner before burning (g) 212.11 213.27 219.85
mass of burner after burning (g) 211.47 212.27 219.25
mass of fuel used (g) 0.64 0.69 0.60
mass of water heated (g) 200.00 200.00 200.00
temperature of water before heating (°C) 18.00 18.00 19.00
temperature of water after heating (°C) 34.00 40.00 43.00
to find out which was the most suitable fuel out of those three using calcs..
eg..for methanol...
C2H5OH (l) + 3O2 -----> 2CO2 (g) + 3H2O (L)
h=mc T
= 200 x 4.2 x 22
= 18.48 kJ
H= h / m/M
= 18.48 / (0.69/46)
= 1246 kJ mol -¹ released...
so would i have to do this for each spirit and then which eva has the least amount of heat released would be the most effiecient?