# 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??
Original post by HighlightOfUrDay
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?
Original post by TypicalNerd
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?

Thank you so much! It won't let me attach a ss of it but I copied the text:

A bomb calorimeter can be used for accurate determination of the heat change
during combustion of a fuel.
A bomb calorimeter is a container of fixed volume that withstands the change in
pressure during the reaction.
The fuel is mixed with pure oxygen in the calorimeter, ignited and the
temperature change is recorded.
The total heat capacity (Ccal) of the calorimeter is calculated using a fuel for
which the heat change is known.
In an experiment to calculate Ccal, 2.00 g of hexane (Mr = 86.0) is ignited. A
temperature change (∆T) of 12.4 °C is recorded.
Under the conditions of the experiment, 1.00 mol of hexane releases 4154 kJ of
energy when combusted.
(a) The heat energy released in the calorimeter, q = Ccal∆T
Calculate the heat capacity (Ccal) in kJ K−1.

(b) When the experiment is repeated with 2.00 g of octane (Mr = 114.0) the
temperature change recorded is 12.2 °C
Calculate the heat change, in kJ mol−1, for octane in this combustion
reaction.
If you were unable to calculate a value for Ccal in part (a), use 6.52 kJ K−1
(this is not the correct value).
Original post by HighlightOfUrDay
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.
(edited 11 months ago)
Original post by TypicalNerd
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!
Original post by HighlightOfUrDay
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.
Original post by TypicalNerd
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.

Yeah I think that's a good idea. My exam technique is definitely improving, I guess this was just a bump along the way. But yeah, I will keep that in mind and see how I do with implementing that in my exams, thank you so so much!!