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
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