# Homework Help: Final temperature of diamond using Cv

1. Jun 2, 2012

### Ryomega

1. The problem statement, all variables and given/known data

At low temperatures, Cv of solids is given by:

Cv = D($\frac{T}{\Theta}$)3

where:

D = 1.94*10^3
$\Theta$ = Debye Temperature, for diamond 1860K

If 100J of heat is transferred to 1mole of diamond, initially at 10K. Calculate the final temperature.

2. Relevant equations

Cv = D($\frac{T}{\Theta}$)3
Q = mCv(Tf-Ti)

3. The attempt at a solution

This is a straight forward question as the equation becomes:

Q = mCv(Tf-Ti)

Tf = $\frac{Q}{mC}$+Ti

So the real problem I'm having is the fact that Cv is a function of T and T is changing as heat is being put in.

2. Jun 2, 2012

### Ryomega

Oh and I'm guessing mass is just 1 molar mass of carbon

3. Jun 2, 2012

### dipole

dQ = CvdT

Plug in your expression for Cv and integrate from 10 to T:

Q = 1000J = ∫CvdT' from [10, T]

then solve for T.

4. Jun 2, 2012

### Ryomega

Oh... that would do it... *Face Palm*

Can't BELIEVE I didn't see that.:uhh:

Thanks!

5. Jun 2, 2012

### Ryomega

Actually how do you propose to do that?

Cv is the only variable changing with Temperature so:

Cv = D$\int$$\frac{T^3}{θ^3}$

Cv = $\frac{D}{θ^3}$$\int$T3 (from 10 to Tf)

Cv = $\frac{D}{4θ^3}$ (T4f-104)

Q =mCv (Tf-10)

Q =$\frac{mD}{4θ^3}$ (T4f-104)

Let A = $\frac{mD}{4θ^3}$

Q =A (Tf-10) (T4f-104)

At this point it becomes pretty much impractical to solve. Am I doing something wrong?

6. Jun 2, 2012

### dipole

Q is the heat you added to the system, 1000J. It's just simple algebra from there.

7. Jun 3, 2012

### Ryomega

If we expand it (Q = 100J btw)

$\frac{T^5}{10}$ -T4-103T-104=$\frac{10}{A}$

I suppose you're going to tell me I have to do long division? No easier way?

8. Jun 3, 2012

### dipole

No, your integration is wrong here.

Q = ∫CvdT = A(T4 -104)

Q/A + 104 = T4

T = (Q/A + 104)1/4

9. Jun 3, 2012

### Ryomega

Wow... I should go back to school. Thanks a lot.