# Chemistry

heat of combustion of propane c3h8, 2220KJ/mol. the specific heat of copper is 0.385 J/g*C. How many grams of propane must be burned to raise the temperature of a 10.0kg block of copper from 25.0 C to 65.0 C,assuming none of the heat is lost to the surroundings

I'm stuck on a part:
q=m*c*(change of T)
q=10000g(0.385J/g*c)*(65.0C-25.0C)or (338.2 K-298.2K)
q=154000KJ
154000KJ*(1 mol/2220KJ)=69.36936mol
here's where I'm stuck
69.369 mol
and i know that for every 1 mol there is 44.11g of C3H8.
I set it up like 69.369 mol* (44.11g C3H8)/1mol = 3059.88g
is this right?

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3. 31
1. Looks right to me

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posted by GanonTEK
2. Check the units. That is where the error is.

q=mct

where

m=10.0kg=10,000g or 10 x 10^3g
c=0.385 J/g*C
and
T=65 C - 25 C=40 C

q=(10,000g)(0.385 J/g*C)(40 C)

q=15.4 x 10^4 J

We agree, but I have 154,000J and not KJ. You have to convert 154,000J to KJ.

154,000J =15.4 x 10^1 KJ= 154KJ

154KJ*(1 mol/2220KJ)=0.06937 moles=6.937 x 10^-2 moles

6.937 x 10^-2 moles*(44.11g C3H8/mol)=3.06g of C3H8

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posted by Devron
3. Devron is right,
Your caluclation was correct just off by a factor of 1000 as a result.
My apologies for not noticing.

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posted by GanonTEK

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