J.R. S. answered 09/22/21
Ph.D. University Professor with 10+ years Tutoring Experience
Question 1:
q = mC∆T
q = heat = ?
m = mass of water = 1000 g
C = specific heat of water = 4.184 J/gº
∆T = change in temperature = 24.3 - 21.2 = 3.1º
q = (1000 g)(4.184 J/gº)(3.1º)
q = 12,970 J of heat released when 1.0 g sample is burned
Question 2:
1 Cal (nutritional) = 1000 cal
4.184 J = 1 cal
Using these conversion factors we can find calories (cal) and nutritional calories (Cal):
12,970 J x 1 cal / 4.184 J = 3100 calories (cal)/g
3100 cal x 1 Cal / 1000 cal = 3.1 Calories (Cal) nutritional calories/g
Question 3:
Total caloric content of the candy = 3.1 Cal/g x 75.0 g = 233 Calories (nutritional)
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It occurred to me that you could use specific heat of water (C) as 1 cal/gº instead of using 4.184 J/gº and then you wouldn't have to do any conversions of J (joules) to calories. The answer will come out the same, just a little less work.