Energetics question

A vessel and its contents of total heat capacity 120 J K–1 were heated using a methane burner. Calculate the maximum theoretical temperature rise when 0.10 g of methane was completely burned. The standard enthalpy of combustion of methane is –890 kJ mol–1. (4 marks)

Heat capacity is the amount of energy needed to change the temperature of something by 1ºC.

How do you propose to continue?

do you work out the q=mcat bit??

No.

You know how much energy 1 mol of methane releases.
You know the mass of methane so you can work out the number of moles.
Hence you can work out the exact energy released.

AND you know how much energy it takes to change the calorimeter by 1ºC

so ...

moles methane = 16/0.10 = 6.25 × 10–3 (1)
kJ evolved = (6.25 × 10^3 ) × 890 = 5.56 (1)
(5.56 × 10^3 joules) = (mc)∆T (1)
∆T = (5.56 ×10^3)/120= 46.4 K (1)

I pretty much get all the steps until the last step. I don't get why you have to divide by 120? I know it says in the question that 120 the total heat capacity of the vessel and its contents but what does that mean exactly?

Your issue is that you don't really understand the concept of heat capacity.

The specific heat capacity is the amount of energy required to change the temperature of 1g of a substance by 1ºC

However, a vessel being heated may be composed of several substances, so it makes sense to find out the actual heat capacity of the vessel (calorimeter). This is the amount of energy required to change the temperature of the whole calorimeter by 1ºC.

So, if the calorimeter has a heat capacity of 120 kJ K^-1 this means that giving it 120kJ causes an increase in temperature of 1ºC.

Give it 240kJ and the temperature changes by 2ºC, etc.

Hence you have the equation:

ΔE = cΔT

So to calculate ΔT you must divide ΔE by 120

moles methane = 16/0.10 = 6.25 × 10–3 (1)
kJ evolved = (6.25 × 10^3 ) × 890 = 5.56 (1)
(5.56 × 10^3 joules) = (mc)∆T (1)
∆T = (5.56 ×10^3)/120= 46.4 K (1)

I pretty much get all the steps until the last step. I don't get why you have to divide by 120? I know it says in the question that 120 the total heat capacity of the vessel and its contents but what does that mean exactly?

hi can I ask why did you change the 5.56 to standard form please
thanks Bethany