In: Chemistry
4. In total, how much heat is required to raise the temperature of a 450 g sample of water (c = 4.18 J/oC.g) in a saucepan that has a heat capacity of 300 J/oC from 200C to 75oC
a. If I burn 0.315 moles of hexane (C6H14) in a bomb calorimeter containing 5.65 liters of water, what’s the molar heat of combustion of hexane is the water temperature rises by 55.40 oC? The heat capacity of water is 4.18 J/g 0C.
b. If I burn 22.0 grams of propane (C3H8) in a bomb calorimeter containing 3.25 liters of water, what’s the molar heat of combustion of propane if the water temperature rises by 29.50 oC?
4.
We know that
q = m C dT
Where q = heat
m = mass
C= specific heat
dT = change in temperature.
So, heat required to change the temperature of water from 20 oC to 75 oC is
q = (450 g) x (4.18 J oC-1 g-1)
(70 oC – 20 oC)
= (450 g) x (4.18 J oC-1
g-1) (50 oC)
= 94050 J
(a)
We know that; Heat lost = Heat gain
For water
Volume = 5.65 L = 5650 mL
Density of water = 1 g/mL
So, mass of water = volume x density
= 5650 mL x 1 g/mL
= 5650 g
Heat capacity of water, C = 4.18 J oC-1 g-1
Change in temperature = 55.4 oC
So, heat gain by water = (5650 g) x (4.18 J
oC-1 g-1) (55.4
oC)
= 1308382 J
Now, heat evolved (or lost) during the combustion of hexane = heat gain by water = 1308382 J
So,
0.315 moles of hexane produce heat = 1308382 J
1 moles of hexane produce heat = (1308382 J / 0.315)
= 4153594 J
= 4153.594 kJ
= 4154 kJ
Hence, molar heat of combustion of hexane = 4154 kJ/mol
(b)
We know that; Heat lost = Heat gain
For water
Volume = 3.25 L = 3250 mL
Density of water = 1 g/mL
So, mass of water = volume x density
= 3250 mL x 1 g/mL
= 3250 g
Heat capacity of water, C = 4.18 J oC-1 g-1
Change in temperature = 29.50 oC
So, heat gain by water = (3250 g) x (4.18 J
oC-1 g-1) (29.50
oC)
= 400757.5 J
Now, heat evolved (or lost) during the combustion of hexane = heat gain by water = 400757.5 J
Now,
Mass of propane (C3H8) = 22 g
Molar mass of propane (C3H8) = 44 g/mol
So, moles of propane = mass / molar mass
= (22 g) / (44 g/mol)
= 0.5 mol
So,
0.5 moles of propane produce heat = 400757.5 J
1 moles of propane produce heat = (400757.5 J / 0.5)
= 801515 J
= 801.515 kJ
= 802 kJ
Hence, molar heat of combustion of propane = 802 kJ/mol