in need of srs chem help - A-level

I have answered this question on energetics, and one of the questions was asking me about heat change?? Idk if it's different to enthalpy change in terms of the equation, but this question didn't use a mass, not even in the MS. The equation given in the MS was just Q=c(delta)T. I'm super confused: why is there no mass or is this something I should know about??

It would help if you could post the question itself.

Was it asking how much energy is required, per gram, for a particular temperature change?

I have answered this question on energetics, and one of the questions was asking me about heat change?? Idk if it's different to enthalpy change in terms of the equation, but this question didn't use a mass, not even in the MS. The equation given in the MS was just Q=c(delta)T. I'm super confused: why is there no mass or is this something I should know about??

Ok now that I have actually seen the question, I know what you mean.

The difference between an “enthalpy change” and a “heat energy change” is that an enthalpy change is how much (heat) energy is released or absorbed under a constant pressure. A “heat energy change” is more or less the same, however, the pressure does not necessarily have to be constant.

In a bomb calorimeter, the pressure isn’t constant. This is implied by the big ugly block of text at the start of the question. This isn’t especially relevant to the calculations however.

As for the questions themselves:

(a): They gave you the equation q = (Ccal)ΔT. This is because you aren’t expected to learn this equation, so don’t worry about it too much.

In the block of text above, it tells you that for every mole of hexane burned, 4154 kJ of energy is released.

As such, q = moles of hexane x 4154

So first, calculate q.

They also tell you that ΔT is 12.4°C. Although the units of Ccal are kJ K^-1, you do not need to add 273 to ΔT because it is a temperature change as opposed to just a temperature. So temperature change of 12.4°C will be a temperature change of 12.4 K, too.

Now use your value of q, the fact that ΔT = 12.4 K and the formula q = (Ccal)ΔT to find Ccal. You can either rearrange the equation or just plug the numbers into the equation as is.

(b) This requires you to use the same formula as you are given in part (a).

This is hinted by the fact it implicitly tells you to use the same value of Ccal as you calculated in (a) (i.e “If you were unable to calculate a value for Ccal in part (a), use 6.52 kJ K^-1”) and you shouldn’t know of any other formulae that have a Ccal in them.

So you just have to use q = (Ccal)ΔT with your answer to part (a) and ΔT = 12.2 (given in the question), then divide your answer by the number of moles of octane burned.

Thank you so much that makes so much sense omg!! I can't type enough thank you's for this! I think I over-complicate things too much and confuse myself and this was definitely one of those cases. But genuinely, you're a life saver!

I think I would agree with that assessment. But now that you’ve identified that weakness, you can at least try to do something about it.

I’d suggest using a highlighter or simply underlining important sentences, words, numbers and equations given in the questions to help you keep track of them.