Michael S. answered • 08/07/26
B.S. in Chemistry, Indiana University; Organic Chem Teaching Intern
Answer: the 5.0 g of Cheetos that burned released 1.50 Cal, which works out to 0.30 Cal/g.
The one idea this problem is built around: a calorimeter never measures the food. It measures the water. So every quantity in q = m c dT belongs to the water, and the Cheetos only show up at the very end, when you divide.
Step 1 - how much heat the water absorbed.
dT = 30.0 degrees C - 15.0 degrees C = 15.0 degrees C q(water) = m x c x dT q(water) = (100.0 g) x (0.001 Cal/g degrees C) x (15.0 degrees C) q(water) = 1.50 Cal
That 1.50 Cal is the food energy released by the portion that burned, and it is what the question asks you to box.
Step 2 - put it on a per-gram basis. Only the mass that actually burned counts, so subtract first:
mass burned = 85.0 g - 80.0 g = 5.0 g energy per gram = 1.50 Cal / 5.0 g = 0.30 Cal/g
Sig figs, since the problem asks for them. The subtraction 30.0 - 15.0 = 15.0 keeps the tenths place, so dT carries three sig figs, and 100.0 g carries four. The 0.001 Cal/g degrees C is not a measurement at all - it is the ordinary 1 cal/g degrees C rewritten in food Calories, since 1 Cal = 1 kcal = 1000 cal - so it is exact and limits nothing. Three sig figs: 1.50 Cal. The mass difference 85.0 - 80.0 = 5.0 g keeps only the tenths place and so lands on two sig figs, which is why the per-gram value is reported as 0.30 Cal/g and not 0.300.
Check it a second way, with the conversion in a different place. Running the same data in joules with the familiar 4.184 puts the Calorie conversion at the end instead of hiding it inside the specific heat, so agreement here is a real check rather than the same chain run twice:
q = (100.0 g) x (4.184 J/g degrees C) x (15.0 degrees C) = 6276 J 6276 J / (4184 J per Cal) = 1.50 Cal
Three wrong answers this problem reliably produces, each one diagnosable:
WRONG (5.0 g) x (0.001) x (15.0) = 0.075 Cal used the Cheetos mass, not the water WRONG 1.50 Cal / 85.0 g = 0.0176 Cal/g used the whole chip, not the part burned WRONG (100.0 g) x (0.001) x (30.0) = 3.00 Cal used the final temperature as dT
The first is by far the commonest, and it is the one worth fixing permanently: heat flowed out of the Cheetos and into the water, so the mass in q = m c dT is always the mass of whatever warmed up. The second uses the whole chip instead of the part that actually burned. The third forgets that dT is a change, not a final reading.
The most interesting part, and worth a sentence in your writeup. A real bag of Hot Cheetos runs about 5 Cal per gram - roughly 160 Cal in a 28 g bag. This experiment recovered 0.30 Cal/g, about six percent of that. The chemistry is not wrong; the calorimeter is leaky. Most of the heat went into the air, the stand and the can rather than the water, and the chip almost certainly did not burn all the way. That gap is the real lesson of the food-burning lab: an open setup like this systematically underreports, which is exactly why the numbers on a nutrition label come from a sealed bomb calorimeter instead.
One note on signs if your teacher wants them: q(water) = +1.50 Cal and q(reaction) = -1.50 Cal, because the burning released precisely the energy the water took in. Food energy is quoted as the positive magnitude, so the boxed answer stays 1.50 Cal.
