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Topic: Specific Heat Problems  (Read 4629 times)

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Offline schmitgreg

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Specific Heat Problems
« on: December 06, 2007, 06:21:18 PM »
I am having trouble understanding specific heat. My teacher isn't that great at explaining it to me, but he gave us this equation:

s = q / m * ΔT
where
s = specific heat
q = energy in joules
ΔT = change in temperature (Kelvin)

I don't understand what specific heat represents, what to put in for the change in temperature or anything! He said something about water having higher specific heat because it takes more joules of energy to heat it up.

Also, I can plug in the numbers, but I have trouble understanding how to work the equation after plugging in the numbers.

Example: If I want to raise the temperature of one gram of water one Kelvin, then how many joules of energy must I put in?

I put in the numbers and get:

4.186 j/g = q / 1g * 1K

From here I am lost!

Can someone clarify this for me, thanks!

Offline Sev

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Re: Specific Heat Problems
« Reply #1 on: December 06, 2007, 06:28:13 PM »
Quote
Example: If I want to raise the temperature of one gram of water one Kelvin, then how many joules of energy must I put in?

Specific heat capacity of water is 4.18 Jg-1K-1 (so it takes 4.18 J to raise 1g of water by 1K).


Offline enahs

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Re: Specific Heat Problems
« Reply #2 on: December 06, 2007, 06:32:23 PM »
It might be a little easier to look at the equation not solving for specific heat (which is generally notated by "C" by the way) but as the energy in the form of heat of the system.

That is:

q = mCΔT

q = energy
m = mass
C = specific heat
ΔT = Change in temperature (Always Final Temperature - Initial Temperature)

The units for specific heat (in energy of Joules) is: J / (g K)
More formally:
 J / (1g 1K)
That is, the specific heat is how much energy is required to raise 1 gram of the substance one Kelvin unit.

For water that is:
4.184 J / ( g K)   [remember, any unit by it's self there is always a 1 implied, so it is 4.184 J / (1g 1K) ]

It takes 4.184 J to raise 1 gram of water 1 Kelvin unit.

As you see in the q=mCΔT equation, the m (g) and ΔT (K) cancel out the denominator of the specific heat and leave you with joules.

« Last Edit: December 06, 2007, 06:50:31 PM by enahs »

Offline schmitgreg

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Re: Specific Heat Problems
« Reply #3 on: December 07, 2007, 12:53:17 PM »
OK, I understand. I was not understanding what specific heat was. Now I get that it is the amount of energy for one gram and one kelvin unit, so you multiply by the chang in temp. and the mass to get basically the change in joules that needs to occur in order to meet the requirements of the change in temp. that was given.

Thanks for the input guys!

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