Question

Suppose a boil water notice is sent out advising all residents in the area to boil their water before drinking it or using it for cooking. You need to boil 14.0 L of water using your natural gas (primarily methane) stove. What volume of natural gas is needed to boil the water if only 12.2% of the heat generated goes towards heating the water. Assume the density of methane is 0.668 g/L, the density of water is 1.00 g/mL, and that the water has an initial temperature of 21.5 °C. Enthalpy of formation values can be found in this table. Assume that gaseous water is formed in the combustion of methane.

Answer 1

Data

Density of methane = 0.668 g/L

Density of water = 1 g/ml

Volume of water = 14 L

Percent of heat = 12.2 %

Initial temperature = 21.5°C

Volume of natural gas = ?

mass of water = density x volume = 1000 x 14.4 = 14400 g or 14.4 kg

C of water = 4.186 J/g°C

Hf° CH₄ = 74.9 kJ/mol

L = 2257 J/g

Process

1.- Write an equation of the energy needed to evaporate water

E = Sensible heat (liquid phase) + Latent heat of vaporization

Sensible heat = mC(T_{2 -} T₁) T₂ = 100°C Sensible heat = 14400 x 4.186 x ( 100 - 21.5)

= 4731854.4 J

Latent heat of vaporization = mL L = 226 kJ/kg Latent heat = 14400 x 2257 J/g = 32500800 J

E = 4731854.4 J + 32500800 J = 37232654.4 J = 37232.65 kJ

Also,

Density of methane = mass/ volume

mass = density x volume mass = 0.668 x V

and 12.2% of methane of total energy of methane = 37232.65 kJ

0.122 x 74.9 kJ/mol x 0.668 g/L x V = 37232.65 kJ

V = 37232.65 / (0.122 x 74.9 x 0.668)

V = 37232.65 / 6.10

V = 6103.7 L

The volume needed is 6103.7 L