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Calorimetry Lab

I'm not going to lie to you here. For this lab, we pretty much lt stuff on fire. And by "pretty much," I mean entirely. However, there was method to this madness. For this expirement, we used a so called "calorimetry," which is a device used for finding the calories/Calories in a specific food substance.

To start, here is the pre-lab document:

From the pre-lab, I began to understand some of the properties of the physics applied to the lab. Afterward, we began the experiment.

The basis of the lab was as follows: take a large coffee can, place it at the bottom a cork with a wooden skewer embeded in it, straight up and down. Over this setup, suspend a soda can full with 50ml of room temperature water. Then, impale a test material food substance on the skewer. Then, ignite it. As violent as this seems, the exothemic reaction generated from the material can warm the water to such a degree so as to accurately gauge the fuel used on behalf of the material. This only works, mind you, if the waters temperature is carefully gauged.

To conclude, I write the answers required for the post lab write up, using the tables of gathered data during the lab.

Mass of can Mass of can plus water Mass of water
16.8g 67.2g 50g
15.3g 65.6g 50.7g
14.7g 64.5g 50.2g
14.9g 65g 50.1g
17g 70.2g 53.4g
15.4g 67.5g 52.3g


Food Initial mass Final mass Δmass Initial temp Final temp ΔTemp
Marshmallow 11g 10g 1g 22°C 30°C 8°C
Marshmallow 2 12g 11.1g 0.9g 22°C 34°C 12°C
Cheeto 5.8g 5.0g 0.8g 22°C 48°C 26°C
Cheeto 2 5.1g 4.7g 0.4g 22°C 36°C 14°C
Spicey Cheeto 5.1g 4.7g 0.4g 22°C 57°C 35°C
Spicy Cheeto 2 4.3g 4.5g 0.2g 22°C 28°C 6°C


1: Calculate how many calories were transferred to the water by the marshmallow.

Q=50g*4.186 joule/gram*8°C

So then, Q equals 1674.4 calories, which is equal to 1.6744 Calories.

2: Divide this number of calories by the number of grams of marshmallow that burned up to give you the number of calories per gram in a marshmallow.

Since the first burn resulted in one gram burnt, it is convienently 1.6744 Calories to every gram.

3: Marshmallows actually contain about 3.33 calories per gram. Based on this, calculate what percent of the heat you lost to the air around the calorimeter.

Well now I feel a bit like an idiot. Was Literally half of the heat lost to atmosphere, or was there 50% human error?

4: Repeat steps 1 through 3 for a Cheeto. Solve the ratio below to figure out the actual calories per gram in a Cheeto.


5441.8(Q)=50g*4.186 joule/gram*26°C

6802.25 calories per gram, or 6.80225 per Calorie

13604.5 calories are in a gram

5: How close does your answer for #4 come to the real value for Cheetos (6 calories/gram) ?

Infuriatingly enough, the unaltered statement is 100% accurate. I hate human error....