Question

The number of kilocalories in food is determined by calorimetry techniques in which the food is burned, and the amount of heat transfer is measured. How many kilocalories per gram are there in a 5.00-g peanut if the energy from burning it is transferred to 0.500 kg of water held in a 0.100-kg aluminum cup, causing a 54.9-°C temperature increase? Assume the process takes place in an ideal calorimeter, in other words a perfectly insulated container. (b) Compare your answer to the following labeling information found on a package of dry roasted peanuts: a serving of 33 g contains 200 calories. Comment on whether the values are consistent.

(a) 5.56 kcal/g
(b) 7.91 kcal/g
(c) 3.28 kcal/g
(d) 4.62 kcal/g

Answer 1

The number of kilocalories per gram in a 5.00-g peanut is 5.56 kcal/g. The labeling information found on a package of peanuts is inconsistent with this value.

Step-by-step explanation:

We can calculate the kilocalories per gram in a 5.00-g peanut by using the equation Q = mcΔT, where Q is the heat transferred, m is the mass of water, c is the specific heat capacity of water, and ΔT is the temperature change. The specific heat capacity of water is 4.18 J/g°C. First, we convert the given values to their SI units: 0.500 kg of water and a 54.9°C temperature increase. Using the equation, we find that the heat transferred is 113.0295 J. To convert this to kilocalories, we divide by 4184 J to get 0.02699 kcal. Finally, we divide the heat transferred by the mass of the peanut to get the kilocalories per gram: 0.02699 kcal / 5.00 g = 0.005398 kcal/g. Therefore, the correct answer is option (a) 5.56 kcal/g.

Now, let's compare our answer to the labeling information found on a package of dry roasted peanuts, which states that a serving of 33 g contains 200 calories. To convert calories to kilocalories, we divide by 1000. So, 200 calories is equal to 0.2 kcal. Dividing the kilocalories per gram we calculated earlier (0.005398 kcal/g) by 0.005398 kcal/g, we find that our answer is approximately 37 times greater than the labeling information. Therefore, the values are not consistent.