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anonymous

  • 5 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?

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  1. anonymous
    • 5 years ago
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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?

  2. anonymous
    • 5 years ago
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    I think so.

  3. anonymous
    • 5 years ago
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    Ok I read that wrong we need to solve for q first

  4. anonymous
    • 5 years ago
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    \[q=(41.50 \frac{KJ}{K})(276.02 K)\] \[q=11454.83 KJ\]

  5. anonymous
    • 5 years ago
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    Do you have a conversion factor for nutritional Calories?

  6. anonymous
    • 5 years ago
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    1kcal=4.184 kj is all what I have.

  7. anonymous
    • 5 years ago
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    \[1 g Candy (\frac{11454.83 KJ}{4.40 g Candy})(\frac{1KCal}{4.184 KJ})=622 KCal\] I think for food it's Kcal but you might want to check on that

  8. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    to go from c to k you add 273.15 so in this case we got 276.02

  10. anonymous
    • 5 years ago
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    Does that help?

  11. anonymous
    • 5 years ago
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    Yes. Thank you!

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