4. In total, how much heat is required to raise the temperature of a 450 g sampl

Question

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?

Explanation / Answer

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

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