Answer : The heat of combustion is, 15642 J
Solution :
Formula used :
[tex]q=m\times c\times \Delta T[/tex]
where,
[tex]q[/tex] = heat of combustion = ?
[tex]m[/tex] = mass of fructose = 1.0 g
[tex]c[/tex] = heat capacity of the calorimteter = [tex]9.90KJ/g^oC=9900J/g^oC[/tex]
conversion used : 1 KJ = 1000 J
[tex]\Delta T[/tex] = change in temperature = [tex]1.58^oC[/tex]
Now put all the given values in the above formula, we get
[tex]q=1.0g\times 9900J/g^oC\times 1.58^oC[/tex]
[tex]q=15642J[/tex]
Therefore, the heat of combustion is, 15642 J
Answer:
q combustion = -15.6 kJ (exothermic)
Explanation:
Bomb calorimeter questions are interesting because we are usually given the heat capacity of the bomb calorimeter not the specific heat capacity (which includes grams in the unit). Therefore we don't actually need to include the mass of the fructose as long as we know how much the temperature of the calorimeter changed (ΔT) and the heat capacity of the calorimeter.
We know that:
-q (combustion) = q (calorimeter) and we have enough information to calculate q (calorimeter):
q (calorimeter) = (Heat Capacity)(Change in Temp.)
⇒ q(cal) = (9.90 kJ/°C)(1.58 °C) = 15.642 kJ
= 15.6 kJ (3 sig figs)
Since -q (combustion) = q (calorimeter), then:
q (combustion) = -15.6 kJ (negative sign simply means heat released)
