How to Find Heat Absorbed and Entropy Change in a Copper Block?

Click For Summary

Homework Help Overview

The problem involves calculating the energy absorbed as heat and the change in entropy for a 2.00 kg block of copper as its temperature is increased from 25°C to 100°C. The specific heat of copper is provided, and the context is rooted in thermodynamics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of heat absorbed using the specific heat formula and express uncertainty regarding the calculation of entropy change due to the non-ideal gas nature of copper and the variable temperature.

Discussion Status

Some participants have successfully calculated the heat absorbed but are seeking clarification on how to approach the entropy change. A suggestion to integrate the equations for entropy change has been made, although there is acknowledgment of potential uncertainty in the approach.

Contextual Notes

Participants note the challenge of applying standard equations for ideal gases to a solid material like copper, highlighting the complexity of the problem due to the non-constant temperature during the heating process.

dancergirlie
Messages
194
Reaction score
0

Homework Statement


Find (a) the energy absorbed as heat and (b) the change in entropy of a 2.00 kg block of copper whose
temperature is increased reversibly from 25°C to 100°C. The specific heat of copper is 386 J/kg K .


Homework Equations



dS=dQ/T
dS=nRln(final volume/initial volume) **for an ideal gas**

The Attempt at a Solution

\

Well I got the first part. The energy is just
dQ=cmdT
which is (386)(2)(75K)= 57900J

however, when i got to the second part, I didn't know what to do since the copper isn't an ideal gas, and it has a non-constant temperature so I can't use the first equation. Any help would be appreciated!
 
Physics news on Phys.org
Entropy Thermodynamics problem

Homework Statement



Find (a) the energy absorbed as heat and (b) the change in entropy of a 2.00 kg block of copper whose
temperature is increased reversibly from 25°C to 100°C. The specific heat of copper is 386 J/kg K .

Homework Equations



dS=dQ/T
dS=nr ln(vf/vi) **for an ideal gas**

The Attempt at a Solution



The first part was relatively easy. I just used the equation:
dQ=cmdT
so, dQ=(386)(2)(75)=57900 J

But for the 2nd part, I don't know what to do because copper is not an ideal gas, ruling out the 2nd equation, and the temperature is not constant, meaning I cannot use the first equation. Any help would be great!
 


dS = dQ/T
dQ = cmdT
dS = cmdT/T
Integrate both sides to get delta(S) = cm*ln(Tf/Ti); be sure to use Kelvin.
Its been a while since I've done thermodynamics, so I might be wrong...though I only applied a bit of mathematical procedure, so it seems as though it would be fine.
 


sounds right to me... thanks for the help!
 

Similar threads

Replies
14
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 7 ·
Replies
7
Views
6K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K