Entropy Change of Touching Blocks That Reach Equillibrium

[Joosh]

Hello, folks! So, I've come across this question on my Physics homework, and I'm not entirely sure how to finish all the parts. I've included the parts I've gotten correct and what I've gone to get those answers.

Use the exact values you enter in previous answer(s) to make later calculation(s).

A block of copper at a temperature of 60°C is placed in contact with a block of aluminum at a temperature of 44°C in an insulated container. As a result of a transfer of 1065 J of energy from the copper to the aluminum, the final equilibrium temperature of the two blocks is 48°C.

(a) What is the approximate change in the entropy of the aluminum block?
(1/T) = ΔS/ΔE
Plugging the numbers in, I get
(1/(273+48)K) = ΔS/1065 J
Solving for ΔS gave me 3.316 J/K, which was correct, according to WebAssign.

(b) What is the approximate change in the entropy of the copper block?
I figured it would be -3.316 J/K since the change in energy of the copper would be -1065 J and the temperature would still be 48°C since it's at equillibrium, but WebAssign seems to disagree with me.

(c) What is the approximate change in the entropy of the Universe?

(d) What is the change in the energy of the Universe?
This one is, rather predictably, 0.

Some possibly useful equations:

• (1/T) = ΔS/ΔE
• S = k*lnΩ
• ħ = k√(k/m)
  
• k = 1.4*10^-23 J/K
• Ω = (q+N-1)!/(q!)(N-1)!

Any help would be extremely appreciated! Thanks in advance!

 

[TSny]

Welcome to PF!

I think you might have "lucked out" on part (a). The temperature of the aluminum block changes from 44 ^oC to 48^oC. The block only reaches 48^oC at the very end of the process. So, 48^o does not very well represent the temperature of the block during the process. What would be a better temperature value to use?

 

[Chestermiller]

It really would be more accurate to calculate the changes in entropy of the two blocks using the average temperatures, rather than the final temperature. The average temperatures are 52 C and 46 C.

Oops. I didn't see TSny's post.

 

[Joosh]

Ah, I see what the issue was. Thanks, guys!

 

[haruspex]

Ah, I see what the issue was. Thanks, guys!

If you go through the differential equations, you will find $\Delta S=\frac{\Delta Q}{\Delta T}\ln(1+\frac{\Delta T}{T_i})$.
For small $\Delta T$ that approximates $\frac{\Delta Q}{T_i}$. Expanding the ln to the second term matches using the average temperature, as TSny and Chester suggest.

 

[TSny]

If you go through the differential equations, you will find $\Delta S=\frac{\Delta Q}{\Delta T}\ln(1+\frac{\Delta T}{T_i})$.
For small $\Delta T$ that approximates $\frac{\Delta Q}{T_i}$. Expanding the ln to the second term matches using the average temperature, as TSny and Chester suggest.

I agree with your expression for $\Delta S$. But I'm not getting that the expansion to second order in $\Delta T/T_i$ corresponds to using the average temperature. Would you mind checking this?

 

[haruspex]

I agree with your expression for $\Delta S$. But I'm not getting that the expansion to second order in $\Delta T/T_i$ corresponds to using the average temperature. Would you mind checking this?

Well, it's not an exact match, but the difference is third order.
$\frac{\Delta T}{T_i+\frac 12\Delta T}=\frac{\Delta T}{T_i}(1+\frac{\Delta T}{2T_i})^{-1}= \frac{\Delta T}{T_i}(1-\frac{\Delta T}{2T_i})+O((\frac{\Delta T}{T_i})^3)=\frac{\Delta T}{T_i}-\frac 12(\frac{\Delta T}{T_i})^2+O((\frac{\Delta T}{T_i})^3)=\ln(1+\frac{\Delta T}{T_i})+O((\frac{\Delta T}{T_i})^3)$

 

[TSny]

Well, it's not an exact match, but the difference is third order.
$\frac{\Delta T}{T_i+\frac 12\Delta T}=\frac{\Delta T}{T_i}(1+\frac{\Delta T}{2T_i})^{-1}= \frac{\Delta T}{T_i}(1-\frac{\Delta T}{2T_i})+O((\frac{\Delta T}{T_i})^3)=\frac{\Delta T}{T_i}-\frac 12(\frac{\Delta T}{T_i})^2+O((\frac{\Delta T}{T_i})^3)=\ln(1+\frac{\Delta T}{T_i})+O((\frac{\Delta T}{T_i})^3)$

OK. That's nice. Thanks.