## anonymous 4 years ago 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?

1. anonymous

$1 g Candy (\frac{41.50 KJ}{4.40 g Candy})(\frac{239 calories}{1 KJ})=?$ Can you solve from there?

2. anonymous

I think so.

3. anonymous

Ok I read that wrong we need to solve for q first

4. anonymous

$q=(41.50 \frac{KJ}{K})(276.02 K)$ $q=11454.83 KJ$

5. anonymous

Do you have a conversion factor for nutritional Calories?

6. anonymous

1kcal=4.184 kj is all what I have.

7. anonymous

$1 g Candy (\frac{11454.83 KJ}{4.40 g Candy})(\frac{1KCal}{4.184 KJ})=622 KCal$ I think for food it's Kcal but you might want to check on that

8. anonymous

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.

9. anonymous

to go from c to k you add 273.15 so in this case we got 276.02

10. anonymous

Does that help?

11. anonymous

Yes. Thank you!