Question

Someone please help me pass gen chem....

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 16.5 L of water using your natural gas (primarily methane) stove. What volume of natural gas is needed to boil the water if only 17.9% 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 20.4 °C. Enthalpy of formation values can be found in this table. Assume that gaseous water is formed in the combustion of methane.​

Answer 1

Answer:

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

hope helped!!

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Answer 2

The volume of natural gas needed to boil the water if only 17.9% of the heat is generated towards heating water is ; 918.70 L

Using the given data :

Volume of water = 16.5 L = 16500 mL

mass of water = 16500 g  ( 16500 mL * 1.00 g/mL )

Density of methane = 0.668 g/L

Density of water = 1.00 g/mL

First step : determine the heat needed to boil the water

Heat required = mass of water * specific heat water * ( Δ T )

                        = 16500 * 4.184 * ( 100 - 20.4 ) =  5495265.6 J

                        = 5495.265 kJ

∴ Heat required to boil water = 5495.265 kJ

next step ; determine the heat given by combustion

heat given by combustion = ( Heat required to boil water) / ( % of heat generated )

Heat given by combustion = ( 5495.265 ) / ( 17.9 % )

                                             = 30699.80 kJ

Enthalpy of methane combustion = 802.5 kJ/mol

∴ moles of methane used = ( 30699.80 ) / ( 802.5 ) = 38.26 mol

next ; determine the mass of methane ( natural gas )

= ( moles of methane used ) * ( molar mass )

= 38.26 * 16.04 g/mol = 613.69 g

Final step : Calculate the volume of natural gas is needed to boil the water

= mass of natural gas / density of methane

= 613.69 g / 0.668 g/L

= 918.70 L

Hence we can conclude that the volume of natural gas needed to boil water if only 17.9% of the heat is  918.70 L .

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