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?

Calories = 1gCandy (41.50KJ / 4.40gCandy) (239calories / 1KJ)

q = (41.50 KJ / K) (276.02 K)
q = 11454.83 KJ

Calories = 1gCandy (11454.83 KJ / 4.40gCandy) * (1 KCal / 4.184 KJ)
Calories = 622 KCal

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JannetPalos JannetPalos · Answer 2 · 2019-08-23T07:59:32-04:00

Answer:

Nutritional calories per gram of candy = \(\frac{2.74\times 10^{3}}{4.40}cal/g=6.23\times 10^{2} nutritional cal/g\)

Explanation:

Total amount of heat released due to combustion of the candy = (\(C_{cal}\times \Delta T\))

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 = \((41.50\times 275.87)kJ=11449 kJ\)

Now, 1 kJ = 0.239 nutritional cal

So, 11449 kJ = \((0.239\times 11449)cal=2.74\times 10^{3}cal\)

hence nutritional calories per gram of candy = \(\frac{2.74\times 10^{3}}{4.40}cal/g=6.23\times 10^{2} nutritional cal/g\)