Enthalpy qs (1 Viewer)

[indeed]

I have two Qs about enthalpy:
- For the enthalpy formula, delta H = -m c delta T, do you still divide it by n (moles) even if the question doesn't state molar enthapy?
- And also, in the same formula relating to the 'm' (mass), if a solid is dissolved in a solvent, is the mass the solvent's mass, or the solvent + solid's mass?

Thank you

 

[wizzkids]

The equation you are referring to is this one - q=mCΔT
where q is the quantity of heat in joules, and m is the mass of the system, C is the specific heat capacity of the system, and ΔT is the temperature change.
How you further process that quantity of heat depends on the question. Read the question carefully.
If, for example, the question is asking for the quantity of heat liberated, then no, you don't divide by n.
On the other hand, if the question is asking you to calculate the enthalpy of combustion in kilojoules per mole, then yes, you would determine "n" the number of moles of fuel that combusted, and divide q/n
Regarding the determination of the mass of the system, it includes all the mass that is in the system, so solvent mass plus solute mass (and anything else that is part of the system).
I hope this helps.

 

[indeed]

The equation you are referring to is this one - q=mCΔT
where q is the quantity of heat in joules, and m is the mass of the system, C is the specific heat capacity of the system, and ΔT is the temperature change.
How you further process that quantity of heat depends on the question. Read the question carefully.
If, for example, the question is asking for the quantity of heat liberated, then no, you don't divide by n.
On the other hand, if the question is asking you to calculate the enthalpy of combustion in kilojoules per mole, then yes, you would determine "n" the number of moles of fuel that combusted, and divide q/n
Regarding the determination of the mass of the system, it includes all the mass that is in the system, so solvent mass plus solute mass (and anything else that is part of the system).
I hope this helps.

Oh ok, thank you