Stuck on Reaction Enthelpy Question

[Lori]

Homework Statement

What mass of methane must be combusted with excess oxygen to generate enough heat to warm 675 g of water by 25.0 C? (assume no heat lost to surroundings)

Homework Equations

Specific heat of H20= 4.184 J/g*C
q = mCT
balanced equation: CH4 + 2O2 -> CO2 + 2H2O

The Attempt at a Solution

I'm kinda stuck here. Here's what i found...

from stoic, found that 675g H20 is 37.5 mols of H20 and that its 18.75 mols of CH4

I also figured out that the energy the water absorbs from the reaction is 70605 Joules from m*C*T

every reaction of 18.75 mols of CH4 creates 16693.125 KJ but we need 70605 Joules. Thus, 70605 Joules - 16693.125 KJ = 53911.875 J needed ? I kinda paused here though cause I am not sure if it's right!

 

[Charles Link]

You are warming 675 g of water (in a pot) with a temperature increase $\Delta T=25^o \, C$. You don't need to create 675 g of water. (The water is present in the pot before the reaction). You should know the heat capacity of water is 1.00 cal/(gram-degree centigrade). This allows you to compute the $\Delta Q$ that you need. What is $\Delta H$ per mole for this methane reaction? Then you can compute the number of moles of methane you need.

 

[Lori]

You are warming 675 g of water (in a pot) with a temperature increase $\Delta T=25^o \, C$. You don't need to create 675 g of water. (The water is present in the pot before the reaction). You should know the heat capacity of water is 1.00 cal/(gram-degree centigrade). This allows you to compute the $\Delta Q$ that you need. What is $\Delta H$ per mole for this methane reaction? Then you can compute the number of moles of methane you need.

Is it right to say that the Q of water (m*C*T) is the heat we need to be generated by the combustion?

 

[Charles Link]

Yes .$Q=mC \, \Delta T$ and $C=1.00 cal/(gram-degree Centigrade)$.

 

[Lori]

Yes .$\Delta Q=mC \Delta T$ and $C=1.00 cal/(gram-degree Centigrade)$.

How come we use C = 1.00 instead of 4.184?

 

[Lori]

How come we use C = 1.00 instead of 4.184?

Oh ! 4.184 J = 1 cal

Ok.. so i found that Q for water is 70.605 Joules and that 1mole CH4/890.3 KJ, using dimensional analysis, i will need 0.079 moles of CH4 which is 1.27 g CH4 needed

 

[Charles Link]

Oh ! 4.184 J = 1 cal

Ok.. so i found that Q for water is 70.605 Joules and that 1mole CH4/890.3 KJ, using dimensional analysis, i will need 0.079 moles of CH4 which is 1.27 g CH4 needed

Editing... (you added something)... I think Q should be 70,605 Joules. (or 70.605 kJ). Otherwise, what you have looks correct. :)

 

[Lori]

Editing... (you added something)... I think Q should be 70,605 Joules. (or 70.605 kJ). Otherwise, what you have looks correct. :)

oops! you're right. I meant kilojoules :p