Question

250 gallons of water (at 25 degrees Celsius and 1 atmospheric pressure) are boiled to power a steam engine. The heat source to boil the water is the combustion of methanol according the the reaction below. 2 CH3OH (liters) + 3 O2 (grams) −→ 2 CO2 (grams) + 4 H2O (liters) Using the enthalpy of the combustion of methanol (at 25 degrees Celsius and 1 atmospheric pressure), how many liters of methanol are required to boil the 250 gallons of water?

Answer 1

ANSWER:

CONCEPT: The energy supplied by combustion of methanol is used up for two processes

1) For raising temperature of water to boiling point. (Energy needed = m X Cp X ΔT)

m = mass of water (1 Gallon of water = 3.79Kg, 250 Gallon = 3.79 X 250 = 947.5Kg = 947500g)

Cp = heat capacity of water = 4.18J/gK

ΔT = temperature difference ( Final temperature - Initial Temperature = 100 - 25 = 75C)

2) Conversion of water to steam. ( Energy needed = Heat of vaporization X mass of water)

Heat of vaporization of water = 2258.33 J/g

Total energy needed = (m X Cp X ΔT) + (Heat of vaporization X mass of water)

Total energy needed = (947500 X 4.18 X 75) + ( 2258.33 X 947500)

Total energy needed = (297041250) + (2139767675) = 2436808925J

STEP II) Energy released by combustion of methanol

2CH₃OH (liters) + 3O₂ (grams) → 2CO₂ (grams) + 4H₂O (liters)   ΔH = -715KJ/mol

from the stiochiometery energy released = 2 X 715 = 1430KJ

It implies two moles of methanol ( = 64g) release 1430KJ of energy

or 80.80 mL ( = 0.0808L) gives 1430KJ = 1430000J of energy.

V = mass/ density = 64/0.792 = 80.80

2436808925J of energy will release by 0.0808/1430000 X 2436808925 = 137.68L