Final temperature of diamond using Cv

[Ryomega]

Homework Statement

At low temperatures, Cv of solids is given by:

C_v = D([itex]\frac{T}{\Theta}[/itex])³

where:

D = 1.94*10^3
[itex]\Theta[/itex] = Debye Temperature, for diamond 1860K

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

Homework Equations

C_v = D([itex]\frac{T}{\Theta}[/itex])³
Q = mC_v(T_f-T_i)

The Attempt at a Solution

This is a straight forward question as the equation becomes:

Q = mC_v(T_f-T_i)

Tf = [itex]\frac{Q}{mC}[/itex]+T_i

So the real problem I'm having is the fact that C_v is a function of T and T is changing as heat is being put in.
So how do I go about this question?

Thanks in advance

 

[Ryomega]

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

 

[dipole]

dQ = C_vdT

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

Q = 1000J = ∫C_vdT' from [10, T]

then solve for T.

 

[Ryomega]

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

Can't BELIEVE I didn't see that.

Thanks!

 

[Ryomega]

Actually how do you propose to do that?

C_v is the only variable changing with Temperature so:

C_v = D[itex]\int[/itex][itex]\frac{T^3}{θ^3}[/itex]

C_v = [itex]\frac{D}{θ^3}[/itex][itex]\int[/itex]T³ (from 10 to T_f)

C_v = [itex]\frac{D}{4θ^3}[/itex] (T⁴_f-10⁴)

Q =mC_v (T_f-10)

Q =[itex]\frac{mD}{4θ^3}[/itex] (T⁴_f-10⁴)

Let A = [itex]\frac{mD}{4θ^3}[/itex]

Q =A (T_f-10) (T⁴_f-10⁴)

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

 

[dipole]

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

 

[Ryomega]

If we expand it (Q = 100J btw)

[itex]\frac{T^5}{10}[/itex] -T⁴-10³T-10⁴=[itex]\frac{10}{A}[/itex]

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

 

[dipole]

Actually how do you propose to do that?

C_v is the only variable changing with Temperature so:

C_v = D[itex]\int[/itex][itex]\frac{T^3}{θ^3}[/itex]

C_v = [itex]\frac{D}{θ^3}[/itex][itex]\int[/itex]T³ (from 10 to T_f)

C_v = [itex]\frac{D}{4θ^3}[/itex] (T⁴_f-10⁴)

Q =mC_v (T_f-10)

Q =[itex]\frac{mD}{4θ^3}[/itex] (T⁴_f-10⁴)

Let A = [itex]\frac{mD}{4θ^3}[/itex]

Q =A (T_f-10) (T⁴_f-10⁴)

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

No, your integration is wrong here.

Q = ∫C_vdT = A(T⁴ -10⁴)

Q/A + 10⁴ = T⁴

T = (Q/A + 10⁴)^1/4

 

[Ryomega]

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