Question #144803

Estimate the volume of methane gas needed to bring the water in a 1l kettle to the boil, considering it takes about 4.18 kj to increase the temperature of 1 l of water by 1 degree celsius and that 1 mol of gas takes a volume of about 24 l at ambient temperature and pressure.

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Right, so I know CH4 is 16.04 g/mol, and water boils at 100 degrees celsius. But Im not sure how to go about working this problem or what type of problem it is?

I imagine the 4.12 kj would be multiplied by 100 to get it to be the amount needed to heat water to boil- but then I don't understand the ambient temperature and pressure.

I know ambient temperature is 25 degrees celsius, so that makes me think that I would have to multiply 25 by 4.18 and subtract it from 418- Like this:

100 (4.18) - 25(4.18) = 313.5 kj/mol needed to get it to a boiling temperature.

but I don't understand the rest, and I'm not sure if I understand the problem at all.

--------

Right, so I know CH4 is 16.04 g/mol, and water boils at 100 degrees celsius. But Im not sure how to go about working this problem or what type of problem it is?

I imagine the 4.12 kj would be multiplied by 100 to get it to be the amount needed to heat water to boil- but then I don't understand the ambient temperature and pressure.

I know ambient temperature is 25 degrees celsius, so that makes me think that I would have to multiply 25 by 4.18 and subtract it from 418- Like this:

100 (4.18) - 25(4.18) = 313.5 kj/mol needed to get it to a boiling temperature.

but I don't understand the rest, and I'm not sure if I understand the problem at all.

Expert's answer

Two data are missing : the heat of combustion of methane and the initial temperature of the cold water. If it is 25°

25°C, the amount of heat Q

Q necessary for heating1kg

1kg water from 25

25°C to 100°

100°C is Q=4.18kJ kg

−1

K

−1

⋅75K⋅1kg=313.5

kJ

Q=4.18kJ kgX−1 KX−1⋅75K⋅1kg=313.5kJ. And now the heat of combustion of methane is necessary for going ahead.

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