# Specific heat capacity of a metal at low temperature

1. Mar 29, 2016

### erisedk

1. The problem statement, all variables and given/known data
The specific heat capacity of a metal at low temperature (T) is given as $C_p = 32\left({\dfrac{T}{400}}\right)^3$ (kJK-1kg-1). A 100g vessel of this metal is to be cooled from 20 K to 4 K by a special refrigerator operating at room temperature (27°C). The amount of work required to cool the vessel is

(A) equal to 0.002 kJ
(B) greater than 0.148 kJ
(C) between 0.148 kJ and 0.028 kJ
(D) less than 0.028 kJ

2. Relevant equations
q = u + w
dq = mCpdT

3. The attempt at a solution
I don't know how to calculate the change in internal energy. I do know how to calculate q.
-dq = mCpdT
$q = -\int\limits_{20}^{4}32\left({\dfrac{T}{400}}\right)^3 dT$
q = 0.002 kJ
(A) is not the answer (and it shouldn't be because I've ignored Δu). How do I find Δu so I can then find the work done from the first law equation? Please help.

2. Mar 29, 2016

### SteamKing

Staff Emeritus
What happened to m in your integral evaluation? Can you show your calculations for the evaluation of the integral?

3. Mar 29, 2016

### erisedk

Oh sorry, I did include it in the calculation. Here:
$q = -(0.1)\int\limits_{20}^{4}32\left({\dfrac{T}{400}}\right)^3 dT$

$= -(0.1)\frac{32}{400^3}.\frac{1}{4} T^4$ (from 20K to 4K)

$= (0.1)\frac{32}{400^3}.\frac{1}{4} (20^4 - 4^4)$

$= 2 × 10^{-3}$ kJ

4. Mar 29, 2016

### rude man

Problem is, you're violating the 2nd law of thermodynamics.
Reconsider, incorporating that law.

5. Mar 29, 2016

### erisedk

I'm sorry, but I really have no clue here.

6. Mar 29, 2016

### erisedk

I agree I am violating the second law, I just don't know how to proceed after that.

7. Mar 29, 2016

### rude man

Assume no increase in entropy, thern what additional equation can you produce to put a bottom on the work needed? (If you don't know what entropy is you're out of luck).

8. Mar 29, 2016

### erisedk

I do know what entropy is, but I seriously don't know what other equation to use.

9. Mar 29, 2016

### rude man

Ok, fact: the best you can do is have the increase in entropy of the high-temperature reservoir = decrease in entropy of your metal. That's the second law here.

Now, think about what's going on: you start with the metal at 20K and end up at 4K, while the high-temperature reservoir is at a constant 27K. This is therefore a continuous process in which differential amounts of heat are removed from the metal at various temperatures (between 20K and 4K) and pushed up to the high-temp reservoir. So you need to come up with the differential expressions for the first and second laws.

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