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?

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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?

Chemistry
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\[1 g Candy (\frac{41.50 KJ}{4.40 g Candy})(\frac{239 calories}{1 KJ})=?\] Can you solve from there?
I think so.
Ok I read that wrong we need to solve for q first

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\[q=(41.50 \frac{KJ}{K})(276.02 K)\] \[q=11454.83 KJ\]
Do you have a conversion factor for nutritional Calories?
1kcal=4.184 kj is all what I have.
\[1 g Candy (\frac{11454.83 KJ}{4.40 g Candy})(\frac{1KCal}{4.184 KJ})=622 KCal\] I think for food it's Kcal but you might want to check on that
Never mind. This is the conversion factor 1 Cal = 1000 cal = 4.184 kJ What is not clear to me is this part : Also note that a temperature change, ΔT, has the same value in degrees Celsius as it does in kelvins. Thus, ΔT = 2.87 °C = 2.87 K.
to go from c to k you add 273.15 so in this case we got 276.02
Does that help?
Yes. Thank you!

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