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 18.0 L of water using your natrual gas (primarily methane) stove. What volume of natural gas is needed to boil the water if only 19.4% of the heat generated goes towards heating the water. Assume the density of methane is 0.660g/L, the density of water is 1.00 g/ml, and that the water has an initial temperature of 22.7 °C. Enthalpy of formation values can be found in this table. Assume that gaseous water is formed in the combustion of methane. volume of methane:

Answer 1

Solution-

From the question the volume of water = 18 L = 18000 mL

Now we can find the mass of water = (volume of water) * (density of water)

mass of water = (18000 mL) * (1.00 g/mL)

mass of water = 18000 g

Now we find the heat required to boil water = (mass of water) * (specific heat water) * (final temperature - initial temperature)

putting the value the heat required to boil water = (18000 g) * (4.184 J/g.oC) * (100 oC - 22.7 oC)

heat required to boil water = 5821617.6 J

heat required to boil water = 5821.62 kJ

The heat given by the combustion = (heat required to boil water) / (percent of heat taken by boiling)

Heat given by combustion = (5821.62 kJ) / (19.4 /100)

Therefore the heat given by combustion = (5821.62 kJ) / (0.194)

Heat given by combustion = 30008.35 kJ

As we know that the enthalpy of combustion of methane = 802.5 kJ/mol

The moles methane used = (Heat given by combustion) / (enthalpy of combustion of methane)

moles methane used = (30008.35 kJ) / (802.5 kJ/mol)

So the moles methane used = 37.39 mol

Now the mass methane = (moles methane used) * (molar mass methane)

The mass methane = (37.39 mol) * (16.04 g/mol)

The mass methane = 599.74 g

Now the volume methane = (mass methane) / (density of methane)

volume methane = (599.7356 g) / (0.660 g/L)

volume methane = 908.69 L