Question

Question #f631a

Answer 1

`"11 g CH"_4`

Explanation:

The first thing you need to do here is to calculate the amount of heat needed to get `"2 kg"` of liquid water at `25^@"C"` to `"2 kg"` of liquid water at `100^@"C"`.

To do that, use the equation

`color(blue)(ul(color(black)(q = m * c * DeltaT)))`

Here

• `q` is the heat gained by the water
• `m` is the mass of the water
• `c` is the specific heat of water, equal to `"4.18 J g"^(-1)""^@"C"^(-1)`
• `DeltaT` is the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

In your case, you have a mass of

`2 color(red)(cancel(color(black)("kg"))) * (10^3color(white)(.)"g")/(1color(red)(cancel(color(black)("kg")))) = 2 * 10^3` `"g"`

and a change in temperature of

`DeltaT = 100^@"C" - 25^@"C" = 75^@"C"`

which means that you'll have

`q = 2 * 10^3 color(red)(cancel(color(black)("g"))) * "4.18 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 75color(red)(cancel(color(black)(""^@"C")))`

`q = 6.27 * 10^2 * overbrace(10^3 color(white)(.)"J")^(color(blue)("= 1 kJ"))`

`q = 6.27 * 10^2` `"kJ"`

Now, you know that the enthalpy change of combustion for methane is equal to

`DeltaH_"comb" = - "890 kJ mol"^(-1)`

This means that when `1` mole of methane undergoes combustion, `"890 kJ"` of heat are being given off, hence the minus sign used for the value of `DeltaH_"comb"`.

You can thus say that in order to give off `6.27 * 10^2` `"kJ"` of heat, you must burn

`6.27 * 10^2 color(red)(cancel(color(black)("kJ"))) * "1 mole CH"_4/(890color(red)(cancel(color(black)("kJ")))) = "0.7045 moles CH"_4`

To convert this to grams, use the molar mass of methane

`0.7045 color(red)(cancel(color(black)("moles CH"_4))) * "16.05 g"/(1color(red)(cancel(color(black)("mole CH"_4)))) = color(darkgreen)(ul(color(black)("11 g")))`

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the mass of water and for the final temperature.