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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.

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  • leaf green style avatar for user Divisha Singh
    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)
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    • leaf red style avatar for user Richard
      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.
      (3 votes)
  • blobby green style avatar for user iank4611
    Hey sup man bro bro!! So uhm what I want to say is uhm so like so uhm yeah you know haha so uhm at I cannot understand why we devide 4.184. It is kind of little bit confusing bro. Thank you bro bro man man bro.
    (0 votes)
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    • leaf red style avatar for user Richard
      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)
  • area 52 yellow style avatar for user oswin.cervantes1551
    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)
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    • leaf red style avatar for user Richard
      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)
  • stelly yellow style avatar for user Tire Hots Peddler
    At , 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)
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    • leaf red style avatar for user Richard
      Yes, the heat given off by the reaction is the same heat which is gained by the water. Jay explains this at .

      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)
  • blobby green style avatar for user Will Lawson
    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)
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    • leaf red style avatar for user Richard
      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 at only 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)
  • blobby green style avatar for user Karim Abdelmalek
    Will the heat thats transferred at constant pressure always be negative? Since the heat is being transferred not absorbed.
    (1 vote)
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  • blobby green style avatar for user Iroh Wisdom
    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)
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  • mr pants teal style avatar for user samirchokshi11
    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)
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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.