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### Course: Physical Chemistry (Essentials) - Class 11 > Unit 6

Lesson 4: Measurement of U and H- Calorimetry# Worked example: Measuring the energy content of foods using soda-can calorimetry

Calorimetry is often used to evaluate the energy content of different foods. In this video, we'll see how data from a soda-can calorimetry experiment can be used to calculate the energy content of a marshmallow in Calories per gram. Created by Jay.

## Want to join the conversation?

- Are there other reasons as to why heat gets lost in an imperfect insulator aside from being lost to the environment? For example, is it possible for heat to be lost in the form of CO2 and H2O from the chemical process of combustion?(3 votes)
- Well with this set up ideally all the energy released from the combustion of the marshmallow is transferred to the water giving it kinetic energy and manifesting as an increase in temperature. However the water isn't the only thing accepting the energy from combustion we also have to consider the walls of the calorimeter, the thermometer, and the air inside the container. All of these objects also absorb the energy and interfere with the calculations. Essentially any other piece of matter other than the water and absorb the energy intended solely for the water and create error.

Hope that helps.(4 votes)

- Hey sup man bro bro!! So uhm what I want to say is uhm so like so uhm yeah you know haha so uhm at5:56I cannot understand why we devide 4.184. It is kind of little bit confusing bro. Thank you bro bro man man bro.(0 votes)
- Jay is converting joules to calories, which is a unit conversion, so it requires dimensional analysis. This requires a conversion factor, specifically 1 cal = 4.184 J. If you begin with joules, you’d want to multiply by the conversion factor of (1 cal/4.184 J) so that the joule units cross cancel and calories is the sole remaining unit.

Hope that helps.(4 votes)

- At the end he wrote Cal, but he said calories. I thought there was a difference of x1000 when writing Cal instead of calories?(1 vote)
- Cal are still calories, but with that implied kilo prefix. So when spoken both units, calories and Calories, would sound the same but of course differ by a factor of 1000. So to avoid this confusion kcal or food calories are used instead of saying simply Calories in a scientific context.

Hope that helps.(2 votes)

- At3:22, he says Qp for the marshmallow is negative because it gave off heat, so delta H is negative, which means it is an exothermic reaction. Can we also think about the heat gained by the water as Qp? So in terms of the water, delta H would be positive? I feel like I am misunderstanding which Q value is most relevant, unless it depends on the question being asked.(1 vote)
- Yes, the heat given off by the reaction is the same heat which is gained by the water. Jay explains this at3:04.

For any calorimetry experiment, we don’t measure the heat of the reaction directly, rather indirectly through the temperature change of the surrounding calorimeter (which includes water most times). The energy given off by the reaction is the same energy which is gained by the calorimeter which has caused its change in temperature. We say that those two quantities, the energy given off by the reaction and the energy gained by the calorimeter, is the same energy and therefore equivalent magnitudes from the law of conversation of energy. Energy is simply being transferred from one object to another.

While the magnitudes are the same, the signs will be different. From the reaction’s perspective it is losing energy so the heat is negative. But from the calorimeter’s perspective it is gaining energy so the heat is positive. And since this experiment is being done with constant pressure (at atmospheric pressure), we can say that the heats are equivalent to enthalpies. So these heats are also changes in enthalpies for the reaction and calorimeter.

Hope that helps.(2 votes)

- When he said 1 calorie = 4.184J/g-C does this conversion factor apply to any specific heat or just water?

Example: If nickels specific heat is 0.444, can I use the conversion factor 1 calorie = 0.444J/g-C?(1 vote)- A calorie is a unit of energy and so can only truly be converted to other units of energy like joules. So the conversion factor of 1 cal = 4.184 J only relates calories or joules. For example, the conversion at6:00only turns joules into calories.

Specific heat capacity is a composite unit formed from an energy unit, a mass unit, and a temperature unit. We can convert this specific heat capacity to a slightly different unit by converting one of the component units.

Your example doesn’t really make sense because we’re beginning with 0.444; a number without a unit. We can’t convert it to another unit if it doesn’t have a unit to begin with.

But if something had a specific heat capacity of 0.444 cal/(g*K), and we wanted to convert it to J/(g*K), we would multiply 0.444 by 4.184 since every calorie is equal to 4.184 joules. Which would come to 1.86 J/(g*K).

Hope that helps.(1 vote)

- Will the heat thats transferred at constant pressure always be negative? Since the heat is being transferred not absorbed.(1 vote)
- A steel casting at a temperature 725K and weighing 35K is quenched in 150kg oil at 275K.If there are no heat losses, determine the change in entropy.The Specific heat (Cp) of steel is 0.88kJ/kg.K and that of oil is 2.5kJ/kg.K(1 vote)
- Do we have to consider the energy that was absorbed by the can itself? Soda cans are aluminum that are metals that can conduct heat.(1 vote)

## Video transcript

- [Instructor] 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 the 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. And we also need the initial
temperature of the water. So we'll call that Ti. Next, we'd 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 and 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
wanna look at the thermometer for the maximum temperature reached. And when we find that maximum temperature we can go ahead and record it. So we have our final temperature and once the marshmallow
with the cork and the pin cools down we can find our final mass. Let's say the initial mass of our marshmallow pin 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 is
equal to mc delta T 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 4.18 joules per
gram degrees Celsius, and delta T is the change in temperature which should be the final temperature minus the initial temperature of the water which is 30.0 minus 25.0 which is equal to 5.0 degrees Celsius. Grams cancels out, degrees
Celsius cancels out and we find that Q is
equal to +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 magnitudes. So we can write Q is equal to, this time we're gonna write a negative since heat was given off by
the burning of the marshmallow, 1.0 times 10 to the third joules. 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, so 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 to indicate the heat transferred
under constant pressure. That's the definition of the
change in enthalpy delta H. So burning the marshmallow gave off energy which is an exothermic reaction, therefore, the sign for
delta H is negative. Finally, let's relate this soda
can calorimetry experiments 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 cork. And the final mass was 6.00 grams, which means we burned 0.08
grams of marshmallows. And 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 there is one calorie with a lowercase C for every 4.184 joules. Joules will cancel out and this gives us 239
calories with a 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, there are 1000 calories
with the lowercase C for every one kilocalorie. And that's gonna give
us 0.239 kilocalories. And 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 0.239 Calories with a capital
C which is a food calorie. To find the answer
content of the marshmallow in calories per gram,
we just need to divide our food calories by how many
grams of marshmallows we use which was 0.08 grams in our combustion. So dividing by 0.08 grams gives us approximately
three calories per gram of marshmallows. And that's useful because let's say we had one
serving size of marshmallows which is about 30 grams. So 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.