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dolomite
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Homework Statement
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)
Homework 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
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!