# Heat capacity question

## Homework Statement

The combustion of toluene has a ΔErxn of −3.91 × 103kJ/mol. When 1.70 g of toluene (C7H8) undergoes combustion in a bomb calorimeter, the temperature rises from 23.36 ∘C to 36.67 ∘C. Find the heat capacity of the bomb calorimeter to three sig-figs.

q = m*c_s*deltaT

## The Attempt at a Solution

ΔErxn of −3.91 × 103kJ/mol
so ΔErxn of .0184mol toluene = 0.0184* −3.91 × 103kJ/mol = 71.944kJ/mol

c_s = q/(m*deltaT) = -71.944kJ/mol / (1.7g*13.31deg)

= 3.180kJ/mol
(i just want to make sure I calculated this correctly before I submit it (mastering physics) thanks

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epenguin
Homework Helper
Gold Member
If you define what everthig means and set out a logical argument, just ONE step at a time, then if there are any errors you would probably detect pany errors yourself - I am too tired now to work out what you must mean.

But a heat capacity of an object is not kJ/mol, it is kJ/degree, do you understand?

Thanks for the help. I think Im having difficulty with difference between heat caacity and specific heat capacity. But here is what I did for the calculation.

First I got moles of toluene: 1.70g (1mol/92.15g/mol) = 0.0184mol toluene
then i multiplied this by the value for combustion of toluene: ΔErxn of −3.91 × 103kJ/mol: 0.0184mol * −3.91 × 103kJ/mol = -71.944kJ

then i reasoned that -71.944J is the amount of heat evolved in the combustion reaction. Its negative so it means exothermic, which is heat given off by the system.

I called this value q, and inputed it into the equation heat, and specific heat capacity:
q = mc delta T

-71.944J = mc delta T

then rearranged to get specific heat capacity of bomb calorimeter to three sig-figs.

c_s = -71.944J / (1.70g toluene * 13.31degC)

c_s = -3.180 = -3.18J/g*deg C

Chestermiller
Mentor
The mass of the calorimeter is not 1.7 g. When they ask for the heat capacity of the calorimeter, what they are asking for is not C. They are asking for the product of the calorimeter mass m and its specific heat capacity C; i.e., they are asking for mC.

Borek
Mentor
Judging from both your posts you should recheck what "m" in "mcΔT" is.

epenguin
Homework Helper
Gold Member
Sorry I have not been able to answer earlier, I have been and am down with flu.
Minor point is I get slightly different figures from you for the Joules produced by the combustion, 72.14 J
I am taking molecular mass as 92.14, the difference with your figure however does not explain the above discrepancy.

Then you say you are confused about the various definitions. I don't know the formal terminology your teachers require, but broadly heat capacity is crudely just the capacity of any object to 'absorb heat', more precisely the amount of energy (KJ) required to raise its temperature by 1°. An 'extensive property'.

Specific heat capacity (in my day it was just called specific heat) is the property of a substance (intensive property), so here heat capacity per unit of mass, be it Kg or mole.
So you couldn't have as in #1 heat capacity of a calorimeter in kJ/mol - you cannot have a mole of calorimeter! Even a gram or kilogram of calorimeter makes little sense, as it is an inhomogeneous object.
See Wikipedia for extensive and intensive properties.

You have calculated the Joules generated by the combustion, and you know the temperature rise that this causes, so you can calculate just from these two numbers the number of joules that raises the temperature of the calorimeter by 1°. That's all, there is no reason to reintroduce the toluene amount as you have, making your answer seriously out.

I must say I am a bit perplexed by the numbers that come out, plus some conceptual doubts about what is really expressed by the measurement value, but this can be taken up if you come back.