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Estimation of the heat capacity of a calorimeter

AKA as the constant of the calorimeter.

This the very first and basic experiment to perform, along with some calculations, on calorimetry. At the end of any set of calorimetry measurements the heat capacity is usually determined for any given substance. Therefore, it is natural to first determine the heat capacity of the calorimeter in which the experiments are to be performed.

The experiment of calorimetry

As you may have already notice, calorimetry is about heat measurements in the form of thermal energy which is translated as changes in temperature. From this, the heat capacity of substance is derived as well.

Perhaps, the most basic lab experiment for calorimetry derived from the well known heat balance equation,

$Q=m C_P \Delta T$        Eq. (01)

is related to an adiabatic container (not allowing the heat exchange through its boundaries). In Eq. (01) $Q$ is the heat, $m$ is the mass of the substance, $C_P$ is the heat capacity at constant pressure and $\Delta T$ is a temperature difference occurring in the substance (which of course produces heat transfer).

The above was just math. So, how to we, physically, conceive the experiment?

Details on the experiment of calorimetry

Consider the following observations to the experiment to be performed,

  • heat losses should be reduced to its minimum. Since you want to measure the heat using Eq. (01) if heat is, somehow, lost to the environment you will never know how much energy was lost and as a consequence any calculations would be erroneous,
  • temperatures should be carefully measured. Temperature is a way of sensing thermal energy so that measurements of all temperature changes should be done carefully. As you can see, temperatures are at the heart of Eq. (01),
  • since heat losses would change temperature measurements, the boundaries of the system should be adiabatic. In other words, the walls of the system should not allow any heat exchange or more practically: you should have a container thermally insulated,
  • you should accept that there are pressure changes in a closed container. However, if you open the container to add or remove substance the inside of the container will never get pressurised so that the pressure of the surroundings and that inside the container are basically the same. In other words: the expleriment is performed at constant pressure. If your experiments are done using a pressurised container and this variables changes, then you are no longer estimating $C_P$,
  • in this experiments the volume of the substance (liquid water in this case) is to be changed so that you can be sure $C_V$ is not involved.

Again, how does the experimental set up looks like?

Fig. 01 Materials to built your most simple calorimeter.

From Fig. 01, you can easily imagine that a thermally insulated cup of coffe would form a suitable adiabatic container. Also, the temperature can be readily measured with a lab bulb thermometer.

The experiment step by step

We have the heat equation and the materials for the experiment. How do we do it?

Follow the steps below keeping in mind the physics previously given,

Step 1.

Since you are looking for the heat realesed or absorbed due to a change in temperature in a substance you will need two known quantities of a known liquid. In other words: you should know the masses ($m_h$ and $m_c$) and heat capacities at constant pressure for the liquid ($C_{Ph}$ and $C_Pc$). The subscripts $h$ and $c$ stand for the hot and cold samples,
Fig. 02 You use the same mass of water. However, both, temperature and heat capacities are to be different. Temperatures $T_h$ and $T_c$ need to be measured and heat capacities $C_{Ph}$ and $C_Pc$ estimated from tables.



Step 2.

Found the above data by measuring or estimating the mass in the lab from the known volume and the water density. Next, using a formula for heat capacity as a function of temperature you can use the measured temperature to estimate this missing data:

Step 3.

You should mixed the two water samples in the calorimeter and observe the temperature change. Well, this is not as simple as that. Before, mixing you should put it in the context of Eq. (01).

The hot water sample will have a heat equation as follows,

$Q_h=m_h C_{Ph}\left( T_{hi}-T{hf} \right)$        Eq. (02)

where $T_{hi}$ is the initial temperature of the hot water and $T_{hf}$ is the final hot water temperature. Of course, $T_{hi}>T_{hf}$ and $Q_h$ is positive indicating loss of heat.

On the other hand, the cold water sample will have a heat equation ans follows,

$Q_c=m_c C_{Pc}\left( T_{ci}-T{cf} \right)$        Eq. (03)

where $T_{ci}$ is the initial temperature of the cold water and $T_{cf}$ is the final cold water temperature. Of course, $T_{ci}<T_{cf}$ and $Q_c$ is negative indicating absorption of heat (from the hot sample).

You should notice that the signs of $Q_h$ and $Q_c$ were stablished here for our own convinience and understanding.

Also, from Eqs. (02-03) temperatures $T_{hi}$ and $T_{ci}$ are known. However, $T_{hf}=T{cf}$ because it is the final temperature after mixing in the calorimeter. Therefore, $Q_h$ and $Q_c$ can now be easily calculated.

Step 4.

At this point, you should have from the experiment that $Q_h \neq Q_c$. So, where does the rest of the energy went?

The answer is quite straight: into the calorimeter.

If we add up both estimated energies per mass unit, the energy taken by the calorimeter comes up,

$\dfrac{Q_h}{m_h}+\dfrac{Q_c}{m_c}=Q^{(m)}_{cal}$        Eq. (04)

where $Q^{(m)}_{cal}$ is given in J/kg if SI units are used. Equation (04) is in fact another heat balance, but in this case for the whole system.

Finally, the heat capacity at constant pressure of the calorimeter can be calculated from Eq. (04) as follows,

$C_{P-cal}=\dfrac{Q^{(m)}_{cal}}{T_{cf}}$        Eq. (05)

In this case, Eq. (05) uses $T_{cf}$ since the thermal energy of the calorimeter was increased by number of temperature units (not decreased).

This is the end. I hope you find useful this post.


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