What volume of natural gas is needed to boil the water

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 16.0L of water using your natural gas (primarily methane) stove. What volume of natural gas is needed to boil the water if only 11.8% 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 22.6 degrees C. Enthalpy of formation values can be found in this table. Assume that gaseous water is formed in the combustion of methane.

Answer 1

I am assuming that we do not actually want to convert the water to water vapor, just that we want to get the water to a temperature of 100°C. Under that assumption, the energy absorbed by the water will only be used to heat, not to cause a state change.

First, we will look at how much energy is needed to heat up the water by using the specific heat formula, q_water = m_water·C_s,water·∆T_water.

m_water: 16.0 L water (1000 mL / 1 L) = 16000 mL water (1 g / 1 mL) = 16000 g water

q_water = (16000 g) · (4.184 J/(g·°C) ) · (100.0°C – 22.6°C) = 5180000 J (1 kJ / 1000 J) = 5180 kJ

Next, we need to determine how much energy is generated by the combustion of methane using Hess’s Law. Note, you also need the enthalpy of formation of CO₂, which is -393.5 kJ/mol

CH_{4 (g)} + O_{2 (g)} → CO_{2 (g)} + 2 H₂O_(g)     ∆H_comb = ?

Flip enthalpy of formation of methane: CH_{4 (g)} → C_(s) + 2 H_{2 (g)}     ∆H = 74.6 kJ/mol

Enthalpy of formation of carbon dioxide: C_(s) + O_{2 (g)} → CO_{2 (g)}    ∆H = -393.5 kJ/mol

Multiply the enthalpy of formation of water vapor by two: 2 H_{2 (g)} + O_{2 (g)} → 2 H₂O_(g)    ∆H = -483.6 kJ/mol

Combine enthalpy values to determine the enthalpy of combustion: CH_{4 (g)} + O_{2 (g)} → CO_{2 (g)} + 2 H₂O_(g)     ∆H_comb = 74.6 + -393.5 + -483.6 kJ/mol = -802.5 kJ/mol

Considering that only 11.8% of the energy released during combustion of CH₄ can be used to heat the water:

-802.5 kJ/mol · 0.118 = -94.7 kJ/mol

Now, using the idea that –q_rxn = q_water, if we need 5180 kJ of energy, we need enough CH₄ to release -5180 kJ to the water. From this and the usual energy of combustion, we can find our moles of CH₄:

-

5180 kJ (1 mol CH₄ / -94.7 kJ) = 54.7 mol CH₄

Now to convert from moles of CH₄ to grams (using molar mass) and then to volume (using density):

54.7 mol CH₄ (16.043 g / 1 mol CH₄) = 878 g CH4 (1 L / 0.668 g CH₄) = 1310 L CH₄