# Thermochemistry combustion problem

1. Nov 25, 2007

### dolomite

1. The problem statement, all variables and given/known data
How many grams of methane (CH4 (g)) must be combusted to heat 1,200 grams of water from 26.0 degrees Celsius to 87.0 degrees Celsius, assuming water is in liquid state as a product and 100% efficiency in heat transfer?

Given data:
mass of H2O = 1.20 kg
specific heat capacity (C_s) of H2O = 4.184 J/(g*K)
ΔT = 61 degrees Celsius

Known data:
specific heat capacity (C_s) of CH4 = 2087 J/(g*K)

2. Relevant equations
q = m * C_s * ΔT

q = energy needed
m = mass of substance
C_s = specific heat capacity
ΔT = change in temperature in Celsius

when heat transfer is 100% efficient, -q=q

-m1 * C_s1 * ΔT1 = m2 * C_s2 * ΔT2

3. The attempt at a solution

As simple as this problem seemed to me, I've been stuck on it all holiday weekend and I've finally decided to look for some help.

I used the equation I gave to find the energy needed for water to be heated from 26 to 87 degrees Celsius.

-1200g * 4.184 J/(g*K) * 61 C = 306,268.8 J

So the methane needs to produce 306.3 kJ of energy. From several websites I found the specific heat capacity of methane to be around 2.087 kJ/(kg*K) and unless I've totally fried my brain this weekend, is 2087 J/(g*K). The only thing left to calculate the mass of methane needed (given that the specific heat capacity is also correct) is the change in temperature of methane. It's not given and I'm not sure how I'm supposed to find it.

I think the final equation should look something like this:

mass of methane = 306,268.8 J / (2087 J * ΔT2)

Any help would be much appreciated!

2. Nov 25, 2007

### eli64

you don't need the specific heat of methane, (not calculating how much methane will be HEATED, methane is being BURNED)

what's needed is the heat of combustion of methane (in kJ/mol) which is a neg value

you know the energy needed to heat the water which is + (as it is absorbing)

qH2O = -qCH4

-qCH4 = n(CH4) * deltaH(comb)

find n change to grams CH4

3. Nov 25, 2007

### dolomite

Well that makes more sense, thanks! I think I've partially got an understanding. From Wikipedia, the heat of combustion is 802 kJ/mol or 50,125 J/g

I'm not 100% sure on how to use the equation you gave me, -qCH4 = n(CH4) * ΔH(comb)

if that means -306,268.8 J / -50,125 J/g = n(CH4) then that would mean there are 6.11 g of CH4 needed to heat the H2O 61 degrees.

The idea makes sense to me but I'm not sure if the numbers do, if 1 gram of CH4 produces roughly 50,000 J of energy and the water needs about 300,000 to be heated then 6 grams makes sense. But that number seems incredibly insignificant to the 1,200 g of H2O.

I still feel like I'm overlooking something relatively simple.

4. Nov 25, 2007

### eli64

just to be clear
-306,268.8 J / -50,125 J/g = g (CH4)

yes, it doesn't seem like alot of methane is needed but that is because the heat of combustion is so high. Also why methane (component of natural gas) is used for heating homes.

5. Nov 25, 2007

### dolomite

Awesome, that helped a lot, thank you . I think I've got this idea on lock-down.

The only problem left with this question for me was that when I looked up the answer it was 5.52 g, instead of my calculated 6.11. Which, if I calculated the energy needed to heat the water 61 degrees correctly, means methane's heat of combustion would actually be 887.7 kJ/mol.

Any idea where I went wrong?

6. Nov 25, 2007

### eli64

You calculated correctly but it depends on whether H2O(g) or H2O(l) is used as the product in calculating heat of combustion

If you calculate deltaH (comb) for methane using Hess's law (sum heat of prod) - (sum heat of reactants) using H2O(g) you get

CH4(g) + 2O2(g) --> CO2(g) + 2H2O(g) deltaH = -802kJ/mol
http://pslc.ws/fire/cellulos/combcals.htm

if you use H2O(l) you get -890.9 kJ/mol (can do this yourself)

Ideally you should use the measured value which Wiki gave for H2O(g) but obviously your question did not use.