Question

A researcher studying the nutritional value of a new candy places a 4.40-gram sample of the candy inside a bomb calorimeter and combusts it in excess oxygen. The observed temperature increase is 2.87 °C. If the heat capacity of the calorimeter is 41.50 kJ·K–1, how many nutritional Calories are there per gram of the candy?

Answer 1

To calculate the nutritional Calories per gram of the candy, we can use the heat released by the combustion of the candy, given a 4.40-gram sample and an observed temperature increase of 2.87 °C. Using the equation q = mCΔT and converting units, we can find the nutritional Calories per gram.

Step-by-step explanation:

To calculate the nutritional Calories per gram of the candy, we first need to calculate the heat released by the combustion of the candy. We can use the equation q = mCΔT, where q is the heat released, m is the mass of the candy sample, C is the heat capacity of the calorimeter, and ΔT is the temperature increase.

We have a 4.40-gram sample of the candy and an observed temperature increase of 2.87 °C. The heat capacity of the calorimeter is given as 41.50 kJ·K–1.

Plugging in the values, we get q = (4.40 g)(41.50 kJ·K–1)(2.87 °C). The units will cancel out, leaving us with the answer in kJ.

Next, to convert kJ to nutritional Calories, we need to multiply by a conversion factor, where 1 nutritional Calorie is equal to 4.184 kJ. So, to find the nutritional Calories per gram, we divide the calculated q value by the mass of the candy sample.

Therefore, the nutritional Calories per gram of the candy can be calculated as follows:

(q value in kJ) / (4.40 g) = nutritional Calories per gram

Answer 2

Nutritional calories per gram of candy =
`(2.74* 10^(3))/(4.40)cal/g=6.23* 10^(2) nutritional cal/g`

Step-by-step explanation:

Total amount of heat released due to combustion of the candy = (
`C_(cal)* \Delta T`)

where
`C_(cal)` is the heat capacity of calorimeter and
`\Delta T` is the increase in temperature.

Here,
`C_(cal)` = 41.50 kJ/K and
`\Delta T` = (273+2.87) K = 275.87 K

So, Total amount of heat released due to combustion of the candy =
`(41.50* 275.87)kJ=11449 kJ`

Now, 1 kJ = 0.239 nutritional cal

So, 11449 kJ =
`(0.239* 11449)cal=2.74* 10^(3)cal`

hence nutritional calories per gram of candy =
`(2.74* 10^(3))/(4.40)cal/g=6.23* 10^(2) nutritional cal/g`