Propane gas (C3H8) is used to heat a tank of water. If the tank contains 200L of water, what mass of propane will be required to raise its temp. from 20 degrees cel to 65 degrees cel (delta Hcomb for propane = -2220kj/mol)
The correct answer is:547g
My answer is: around 750g
Some please explain how you got 547g as your answer in steps. Thanks!
The correct answer is:547g
My answer is: around 750g
Some please explain how you got 547g as your answer in steps. Thanks!
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200L water = 200000ml water = 200000g water since density = 1g/ml
the molar heat of combustion for propane = 2.2x10^3 but, close enough for now
heat = mass water x specific heat water x ∆T
heat = 2x10^5g x 4.184J/g-ºC x 45ºC
heat = 37656000J or 3.76x10^7J = 3.76x10^4kJ
3.76x10^4kJ / 2220kJ/mole = 16.96moles
16.96moles x 44g/mole = 746.24g
ok, for a mass of 547g, this would be 12.43moles propane
12.43moles propane x 2220kJ/mole = 2.76x10^4kJ total heat released
2.76x10^4kJ = 2.76x10^7J = 2x10^5g x 4.184J/g-ºC x ∆T
∆T would = 33ºC
therefore, 750 is the correct answer unless there is som caveat to this problem
the molar heat of combustion for propane = 2.2x10^3 but, close enough for now
heat = mass water x specific heat water x ∆T
heat = 2x10^5g x 4.184J/g-ºC x 45ºC
heat = 37656000J or 3.76x10^7J = 3.76x10^4kJ
3.76x10^4kJ / 2220kJ/mole = 16.96moles
16.96moles x 44g/mole = 746.24g
ok, for a mass of 547g, this would be 12.43moles propane
12.43moles propane x 2220kJ/mole = 2.76x10^4kJ total heat released
2.76x10^4kJ = 2.76x10^7J = 2x10^5g x 4.184J/g-ºC x ∆T
∆T would = 33ºC
therefore, 750 is the correct answer unless there is som caveat to this problem