Winter 2007 midterm question 2

[704278444]

Is the calculation of question 2 for heat capacity correct?

q= nC delta T
C= q/ (n delta T)

But the answer key gives:
C= (n q)/ delta T

I'm a bit confused! Thanks!

[Sandra]

If i remember correctly, it is because for calorimeter q is different since the mass of the calorimeter is not taken into account. it is q(cal)=C(cal)delta T

But we multiply it by the n of benzoic acid but since it is in grams we change it to moles. Because we need to take into account the substance. I think that may be why. That's how I went about it.

[104277942]

Sandra is right; you're supposed to do the q=C deltaT formula because we don't know the grams of the calorimeter.
You know the delta T= 2.265 degrees C.
Also, you have the enthalpy of combustion of benzoic acid and you need to know how much heat was given off to the calorimeter.
And conceptually you know that heat capacity is kJ or J/degrees Celsius so you multiply moles of benzoic acid by its enthalpy of combustion to cancel out mols and you take that value of kJ divided by the temp change.
The calorimeter gains heat from the benzoic acid losing heat, so that's where the negative sign in front of the mols of benzoic acid comes from.
Even if you forgot that negative sign, you know heat capacity is always positive, so if you end up with a negative heat capacity, go back and look at what may have gone wrong.
Hopefully this made sense!