U3AOS1 Topic 4: Calorimetry

Calorimetry is the method in which heat energy released or absorbed is experimentally measured via a device called a calorimeter

This is usually to test the energy content of food and the energy released by fuels such as ethanol.

Does calorimetry sound oddly familiar to the everyday word - calorie?

Calorie is just the unit of energy required for a fuel to increase the temperature of $1$ g of water by $1$ $\text{°C} $ or $\text{K}$ at SLC

Basically calorie = specific heat capacity of water at SLC (Standard Lab Conditions)

Thus approx. $4.18$ $ \text{J} $ per calorie

[Insert calorimeter diagram]

As per the law of energy conservation, the energy released or absorbed must be gained or lost from somewhere

That’s where water comes in. Temperature change is determined using water

\[ \text{q} = \text{m} \times \text{c} \times \Delta \text{T} \],

$ \text{q} $ is heat $ \text{(J)} $ , $ \text{m} $ is mass $ \text{(g)} $, $ \text{T} $ is change in temperature ($ \text{°C} $ or $ \text{K} $)

\[ \Delta \text{T} = \text{T_final} - \text{T_initial} \]

​

There are two types of calorimeters:

The energy content of food can be determined by relating the energy calculated to the mass of food combusted.

\[ \text{energy\_content} = \frac{\text{energy absorbed by water}}{\text{mass change of food}} \]

However, calorimeters are not $100$ $\%$ efficient , there is always going to be some heat gained or lost from the external environment and not every calorimeter is the same. Some may be in different conditions or models than others.

So how do we account for that?

We calibrate the calorimeter which accounts for the inaccuracies involved with calorimetry from the heat loss or gain from environment

Let's say we input a known $ 200 $ $\text{J}$ of energy, the temperature of the water measured by the thermometer increases by $5$ $\text{°C}$

How do we build a relationship off that?

Well we have to find how many J per degree change in temperature

\[ \frac{ 200 \text{J}}{ 5 \text{°C}} = 40 \frac{\text{J}}{\text{°C}} \]

Therefore for every $40$ $\text{J}$ inputted, the temperature of the water increases by $1$ $\text{°C}$

This relationship is called the calibration factor (CF) and is specific to each individual calorimeter to reduce inaccuracies

\[ \text{CF} = \frac{\text{q}}{\Delta \text{T}_{\text{during calibration}}} \]

Now how do we input a known amount of energy?

We primarily use a heat rod with energy inputted controlled by the voltage, current and time

\[ \text{E}= \text{V} \times \text{I} \times \text{t} \]

Note: $ \text{q}$ means heat whilst $\text{E}$ means energy, as heat is a type of energy both can be substituted for each other

Treat \[ \text{q} = \text{E}_q = \text{E} \]

Or we use a known amount of high purity compound such as benzoic acid

\[ \text{q} =\Delta \text{H} \times \text{n} \],

where $ \text{H}$ is change in enthalpy of high purity compound and $\text{n}$ is number of moles of the compound

Hence substituting each formula into the $\text{CF}$ formula we get:

\[ \text{CF} = \frac{ \text{V} \times \text{I} \times \text{t}}{\Delta \text{T}_{\text{during calibration}}} \] ​​​and \[ \text{CF} = \frac{ \Delta \text{H} \times \text{n}}{\Delta \text{T}_{\text{during calibration}}} \]

Even after we input a known amount of energy, this still doesn’t account for the inaccuracies caused by the environment

We combat this tribulation by graphing the reaction process and extrapolating the results using temperature - time graphs

[insert temperature - time graph perfect + imperfect]

So what do we do after we find the calibration factor?

Now after calibrating, we can test the fuel source. If the fuel source results in a certain temperature change of the water, we can relate that to the energy released

Eg. If after calibration we found that $100$ $\text{J}$ results in a temperature increase of water of $1$ $\text{°C}$, after we burn $2$ $\text{g}$ of muffins results in a temperature increase of $10$ $\text{°C}$, then we know each gram results in a temperature increase of $5$ $\text{°C}$. Thus corresponding to $500$ $\text{J}$ released from burning muffins. 

Here is a formula to formalise the above:

\[ \text{E}_{\text{released from fuel}} = \text{CF} \times \Delta \text{T}_{\text{after calibration}} \]

Note: This is where many students often get confused

The T value from calibration is different to T value after testing fuel

I like to think of it as two separate steps

Calibration (using $ \Delta \text{T} $ during calibration)

Testing fuels (using $ \Delta \text{T} $ after calibration)

This is all to ultimately test the energy content of the food

\[ \Delta \text{H}_{\text{food}} = \frac{ \text{E}_{\text{released from fuel}} }{ \text{m}_{\text{food}} } \]

As calorimetry is very practical based and is just an experiment, VCAA loves to test this topic in an experimental based approach

For example they like to ask:

What are some ways to increase the accuracy of a calorimeter?

1. Use a tight lid
2. Improve insulation around calorimeter
3. Use digital thermometer
4. Use more pure benzoic acid

Normal tip: Just brainstorm as many ways as possible to limit heat loss

You must also know the cause of overestimation and underestimation of heat content of fuel

Overestimation:

1. Less water than required
2. Non-homogeneous fuel mixture

Special tip: to brainstorm overestimation causes easier, link to what changes would result in a greater change in temperature of water

Reasoning behind this is:

\[ \Delta \text{H}_{\text{food}} = \frac{\text{E}_{\text{released from fuel}}}{\text{m}_{\text{food}}} \]

\[ \Delta \text{H}_{\text{food}} = \frac{\text{CF} \times \Delta \text{T}_{\text{after calibration}}}{\text{m}_{\text{food}}} \]

\[ \therefore \Delta \text{H}_{\text{food}} \propto \Delta \text{T}_{\text{after calibration}} \]

Underestimation:

1. Loose lid
2. Analogue thermometer
3. Poor insulation material

We want lower energy content hence corresponding to lower change in temperature of water after combustion of fuel.

Courtesy of Javalab – Specific Heat Simulation