00:01
In this case, what will happen is with increase in number of carbon atoms, enthalpy of combustion is increasing.
00:07
So we can say that increase in number of c.
00:15
That means carbon atoms, this will imply enthalpy of combustion increases.
00:30
Then we can say that in the b part, more importantly, as number of total molecules being generated, being generated from combustion increases.
00:57
This will imply that the enthalpy of combustion, enthalpy of combustion is going to increase.
01:09
So when we talk about the combustion of butane, combustion of butane.
01:15
So, butane is c4h10, right? so c4h10 plus 13 by 2 oxygen is going to form 4 times co2.
01:24
2 plus 5 times h2o and the combustion of hexane when we talk about the hexane combustion hexane is c6 h 14 so this will be 19 by 2 oxygen giving 6 co2 plus 7 h2o here we know the formula that heat is changed is going to be mc delta t where q is the heat transferred m is our mass c our specific heat capacity and delta t is the temperature difference so for c value for water we have the value is equal to 4 .2 june per gram degree celsius therefore we know the density formula that is mass by volume right so since we need to calculate mass it is going to be equal to density times volume that means for density of water it is one gram per m l into one to one of water is present, therefore mass will be equal to 120 gram.
02:31
Now we have the mass, we have the specific heat value, and the temperature change is given to us, or we can say we can calculate delta t is equal to t2 minus t1, which is equal to 45 minus 0 .8.
02:44
Here we have, first of all, t2, t1 is given as 93 .1, and for t2 is given as 45 .8.
02:50
So we can say 45 .8 minus 93 .1, and also the volume which we are given is 100 ml, not 120.
02:58
So we can just make this correction.
03:00
So this becomes 100 gram.
03:02
And now 45 minus 45 .8 minus 93 .1 will give us the value as negative 47 .3 as the delta t value...