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I've been struggling with this problem that I can't seem to figure out. I think I know how to solve it but I think there's missing information.

The combustion of 1 mole of glucose $\ce{C6H12O6}$ releases $\pu{2.82\times10^3 kJ}$ of heat. If $\pu{1.25 g}$ of glucose are burnt in a calorimeter containing $\pu{0.95 kg}$ of water and the temperature of the entire system raises from $\pu{20.10 ^\circ C}$ to $\pu{23.25 ^\circ C}$. What is the heat capacity of the calorimeter?

I think I need the specific heat of glucose (which I haven't found yet), but also I don't know why they give me the heat released by 1 mole of glucose. I need to know the method of how to solve such problems.

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  • $\begingroup$ please give your comment on the accuracy of the measurement of the heat capacity of the calorimeter. explain any assumptions used calorimeter heat capacity in the experiment? and why a cooling curve (T versus t) was used to determine the mixing temperature? and do you expect your value for the specific heat to be too high or low?why? and what is your unknown metal? $\endgroup$
    – user821
    Commented Nov 7, 2012 at 10:58

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$12.5\ \mathrm{kJ}$ of heat was absorbed by the surroundings.

I found this by using the mcat formula and the specific heat capacity of water (4.18 J/(g °C)):

$Q=m\cdot c\cdot\Delta T$

$Q = 950\ \mathrm{g} \times (4.18\ \mathrm{J\cdot g^{-1}\cdot {^\circ C^{-1}}}) \times (23.25\ \mathrm{^\circ C} - 20.10\ \mathrm{^\circ C}) = 12508.7\ \mathrm{J}$

If you wanted to use this whole formula for solving the calorimeter's specific heat capacity, you would need to know the mass of the calorimeter as well, which is not given.

What your book is probably asking is for what is called the "calorimeter constant". This is given in units of $\pu{J/^\circ C}$ notice that it does not include mass.

Note: Sometimes "the calorimeter's specific heat capcity" is used instead of referring to the calorimeter constant, but in this case we cannot find a value which will include mass in the units, so I think it is more clear to use the term "calorimeter constant."

You can determine the constant by this formula: $$Q_\text{cal} = C_\text{cal} \times \Delta T_\text{cal}$$

Where $Q_\text{cal}$ is the energy absorbed, $C$ is the constant and $\Delta T$ is the same as the change in temperature of the water.

You may calculate $Q_\text{cal}$ by using this formula: $$Q_\text{cal} = -(Q_\text{water} + Q_\text{glucose})$$

It may also help to think of $Q_\text{water}$ = $Q_\text{surroundings}$ and $Q_\text{glucose}$ = $Q_\text{system}$

To find $Q_\text{glucose}$ I did: (glucose has lost energy, it is negative value)

$-2820\ \mathrm{kJ} \times 0.007\ \mathrm{mol}$ and $Q_\text{water}$ is simply the $12508.7\ \mathrm{J}$ positive because $\Delta T$ is positive for the surroundings (the system/glucose lost energy)

$Q_\text{cal} = -(12508.7\ \mathrm{J} + (-19740\ \mathrm{J}))$

So my final answer is then: $2.3\times10^3\ \mathrm{J/{^\circ C}}$

It is important that heat capacities are positive, think about what it would mean if this were a negative value.

In the laboratory, it is necessary to do a calculation such as this one before using a calorimeter for anything. Normally it can be done by heating a piece of nickel or something, recording the temperature of the metal and the water, and then dropping the metal into the calorimeter to find the final temperatures, and then calculate the calorimeter constant. You can then proceed with further experimentation using that calorimeter, but only after this constant has been found can you find the specific heat capacity of other materials.

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  • $\begingroup$ First of all, I have to say a massive thank you, because I wouldn't have thought this that way, (specially because I'm quite bit confused about the specific heat-specific heat capacity thing). Second, I'm guessing, if heat capacities were negative, wouldn't it be violating laws of thermodynamics? $\endgroup$
    – ChairOTP
    Commented Sep 7, 2012 at 16:43
  • $\begingroup$ I just found out there are actually systems where the heat capacities are negative, and even though it goes beyond my knowledge, I can't think of how would it be possible, increasing the temperature by losing energy, seems irrational to me. $\endgroup$
    – ChairOTP
    Commented Sep 7, 2012 at 16:57
  • $\begingroup$ I think that is some in-depth physics topic about systems. A system might be able to present the idea of a negative heat capacity, but try to stay focused on actual materials in this case. If a material had a negative heat capacity it would do the opposite of what boiling water does. If water was -4.18 then to raise water by 1 degree celsius, you would need to extract energy stored in the molecules, so putting water in the fridge would cause it to boil in a sense. This is impossible. Like putting water in the stove and waiting for it to freeze... $\endgroup$
    – Leonardo
    Commented Sep 7, 2012 at 23:04
  • $\begingroup$ You missed a very critical assumption. The problem assumes that the heat capacity of the calorimeter is only due to the water which it contains. For a good calorimeter that is mostly true, but not entirely true. $\endgroup$
    – MaxW
    Commented Feb 25, 2018 at 8:18

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