 So one of the fuels that I expect that you will use is ethanol here are some values The mass loss of the spirit burner was three point seven grams So most of you will I am assuming carry out this experiment with some little spirit burners That will contain the alcohol in them and will have a wick that you can light and therefore burn That fuel in order to produce energy and heat the water You'll obviously very importantly need to measure the mass of the burner mass before and The mass after and therefore the mass before minus the mass after gives us the change in mass Now as I mentioned earlier You will get a value of mass here and the temptation is to put it into the MCAT formula And we need to be very careful that we don't do that So our first step we're going to make sure is really clear that this is about the water and nothing else For this value to be about the water We need to make sure that we have a value for mass which if we're heating a hundred grams That's what's going to go there and I'm going to put a hundred because I'm going to use my Q value of four point one eight I'll get exactly the same value if I change that to zero point one kilograms and multiply it by four point one eight times ten to the three For kilograms, but the numbers will be the same and then I'm going to multiply that by the Delta T which in this case you can see was a temperature rise for the water of 54 degrees now of course, I've put it here in degrees C and you'll also find that the Temperature unit that you're given usually is in Kelvin Now, of course you can convert the initial and final temperatures into temperatures in Kelvin But the one thing that we do know is that the Kelvin scale and the Celsius scale both move in Equal steps So therefore if we're talking about a change in temperature then a 54 degree 54 degrees C change in temperature is going to be equal to 54 Kelvin The absolute values will be different, but the change will not so once we have the change in temperature in a degree C We can easily just convert that into Kelvin and so that will go into our equation And when we put those values into our equation, we'll get a certain number Just to recap on that the calculation we had our mass which was a hundred We multiplied that by four point one eight, which was our specific heat and then by fifty four And when we do that we get a number of two two five Seven two jewels Now if I've run straight to the Delta H value here One of the things that we we do need to realize is that when we're talking about Delta H And not the Q value that we're actually going to have a negative value. This is an exothermic reaction The heat's actually been absorbed by the water So the heat's gone into the water and it therefore it's gone out from the reaction the combustion reaction That's occurred So this is actually a negative value for the Delta H because it's an exothermic reaction now at this stage It's probably reasonable for you to start to express your number as Kilojoules because these numbers do tend to get quite large and so more than reasonable for you to convert that number into kilojoules But and this is the important but this is where we need to take into account how long We were burning the fuel so Obviously this number could be any number and is almost a meaningless number if we don't Identify exactly how much fuel was burnt in order to have the temperature of 100 Mills of water to rise by 54 degrees Now we know when that happened that we used 3.7 grams of fuel So what we need to be aware of the fact is that if we standardize this we can actually get the Value of the energy generated a per gram of fuel So the simple way to do that is to take two two point five seven two and Divide that by three point seven and when we do that we get a value of Of course, it's still a negative so minus six point two kilojoules per gram Now again now this is this is a useful unit because it now allows us to compare the amount of energy That's being released in the combustion of different alcohols We can compare each one per gram So if we have a gram of each of these fuels and we would have to Combust one gram of those fuels. What are the different amounts of energy that we could get? But being chemists Grams isn't the term that isn't the unit that we usually use when we're making comparisons between different types of compounds The better one and the more common one that we choose is the mole