A gas company in Massachusetts charges $2.80 for 15.0 ft3 of natural gas (CH4) measured at 20.0°C and 1.00 atm. Calculate the cost of heating 2.00 × 102 mL of water (enough to make a cup of coffee or tea) from 20.0°C to 100.0°C. Assume that only 50.0% of the heat generated by the combustion is used to heat the water; the rest of the heat is lost to the surroundings. Assume that the products of the combustion of methane are CO2(g) and H2O(l).

$0.0238

Explanation:

The energy you need to increase the temperature of water from 20°C to 100°C is obtained from:

Q = C×m×ΔT

Where Q is the energy, C is specific heat of water (4.184J/g°C), m is mass of water (2.00x10²g - Density of water 1g/mL), ΔT is change in temperature (100.0°C - 20.0°C)

Replacing:

Q = 4.184J/g°C × 2.00x10²g × 80.0°C

Q = 66944J = 66.944kJ

As you are assuming the energy of combustion will be just 50.0% to heat the water the energy you need is 66.944kJ × 2 = 133.888kJ

The combustion of methane is:

CH4(g) + 2O2(g) ⟶ CO2(g) + 2H2O(l) ΔH = −890.8kJ

That means 1 mole of methane produce 890.8kJ. As you need 133.888kJ, moles of methane are:

133.888kJ × (1 mol CH₄ / 890.8kJ) = 0.150 moles of CH₄.

Using PV = nRT, moles of 15.0ft³ (424.8L) at 20.0°C (293.15K) and 1.00atm:

1.00atmₓ424.8L = moles CH₄ₓ0.082atmL/molKₓ293.15K

17.67 = moles CH₄

As 17.67 moles of CH₄ cost $2.80, the cost of 0.150 moles of CH₄ is:

0.150 moles CH₄ ₓ ($2.80 / 17.67 moles) =

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.