Worked example: Measuring the energy content of foods using soda-can calorimetry | Khan Academy

Calorimetry refers to the measurement of heat flow, and in this worked example, we're going to burn a marshmallow and find the energy content of the marshmallow. First, let's look at this setup for our soda can calorimeter.

So, our soda can has some water in it. So here's the water in our soda can, and then we also have a thermometer in the soda can to measure the change in the temperature of the water. If you take a stir rod and you put the stir rod through the tab on the soda can, you can attach the soda can to a stand. Next, we can put our marshmallow on a pin that's attached to a piece of cork. Before we start the experiment, we need to take the mass of the marshmallow with the cork and the pin, so we'll call that the initial mass.

We also need the initial temperature of the water, so we'll call that T sub i. Next, we light the marshmallow on fire. As the marshmallow burns, heat is given off, and that heat is transferred to the water in the soda can. Therefore, the water in the soda can will increase in temperature, which we can see on the thermometer. After the marshmallow burns for a little while, we can stop the burning process, and once we stop that, we want to look at the thermometer for the maximum temperature reached.

When we find that maximum temperature, we can go ahead and record it, so we have a final temperature. Once the marshmallow with the cork and the pin cools down, we can find a final mass. Let's say the initial mass of our marshmallow, pin, and cork was 6.08 grams, and the final mass was 6.00 grams. The initial temperature of the water in the soda can was 25.0 degrees Celsius, and the final temperature was 30.0 degrees Celsius.

Also, let's say that we started with 50.0 grams of water in the soda can. Let's calculate the heat gained by the water. To do that, we can use the q = mcΔT equation, where q is equal to the heat transferred, m is the mass of the water, which is 50.0 grams, c is the specific heat of water, which is 4.1 joules per gram degree Celsius, and ΔT is the change in temperature, which would be the final temperature minus the initial temperature of the water, which is 30.0 - 25.0, which is equal to 5.0 degrees Celsius.

Grams cancel out, degrees Celsius cancel out, and we find that q is equal to positive 1.0 times 10 to the third joules. That's to two significant figures. The positive sign here means that heat was gained by the water, which is why we saw an increase in the temperature. If we assume a perfect transfer of heat, so all the heat that was given off by the burning of the marshmallow was transferred to the water, if we think about q for the burning of the marshmallow, it should be equal in magnitude.

So, we can write q = -1.0 times 10 to the third joules since heat was given off by the burning of the marshmallow. So, assuming a perfect transfer of heat, the magnitude of these two numbers is equal. However, there's no way that all of the heat was transferred from the combustion of the marshmallow to the soda can with our simple setup, so definitely not all of it was transferred.

For example, some of it could have been lost simply to the environment, and since the soda can is open to the atmospheric pressure of the environment, I’ll go ahead and write atmospheric pressure in here. This soda can calorimeter is an example of constant pressure calorimetry, and since this is the heat that’s transferred under constant pressure, I can go ahead and write qp here in it to indicate the heat transferred under constant pressure. That's the definition of the change in enthalpy, ΔH.

So, burning the marshmallow gave off energy, which is an exothermic reaction; therefore, the sign for ΔH is negative. Finally, let's relate this soda can calorimetry experiment to calories in everyday life, and so let's find the energy content of the marshmallow in calories per gram. A food calorie has a capital C and in chemistry, there's also a unit of energy with calorie with a lowercase c. So, one food calorie with a capital C is equal to one kilocalorie or 1000 calories with a lowercase c.

When we burned the marshmallow, we started off with a mass of 6.08 grams for the marshmallow, pin, and cork, and the final mass was 6.00 grams, which means we burned 0.08 grams of marshmallows. When we burned the marshmallow, we found there was a transfer of 1.0 times 10 to the third joules of energy. So first let's convert that into calories with a lowercase c.

So, if we multiply by the conversion factor of one calorie with a lowercase c for every 4.184 joules, joules will cancel out, and this gives us 239 calories with the lowercase c. Next, let's convert 239 calories into food calories. So, first let's take 239 calories, and we can multiply by the conversion factor of 1000 calories with the lowercase c for every one kilocalorie, and that's going to give us 0.239 kilocalories.

Going back up here to our chart, remember, one kilocalorie is equal to one food calorie with a capital C. Therefore, we have 0.239 kilocalories, or 0.23 calories with a capital C, which is a food calorie. To find the energy content of the marshmallow in calories per gram, we just need to divide our food calories by how many grams of marshmallows we used, which was 0.08 grams in our combustion.

So, dividing by 0.08 grams gives us approximately three calories per gram of marshmallows. That's useful because let's say we had one serving size of marshmallows, which is about 30 grams. Therefore, if we know the energy content is three calories per gram, we can simply multiply our four marshmallows approximately by three calories per gram, and grams would cancel out, and we would find that those four marshmallows that we wanted to eat are about 90 food calories.