Calculating the change in entropy, did I calculate correctly?

  • Thread starter Thread starter PhyIsOhSoHard
  • Start date Start date
  • Tags Tags
    Change Entropy
PhyIsOhSoHard
Messages
157
Reaction score
0

Homework Statement


A 3.5 kg block of cobber at 100 degrees celsius (373 K) is put in 0.8 kg water at 0 degrees celsius (273 K).
The equilibrium temperature is 30 degrees celsius (303 K).
Calculate the change of entropy for the system of cobber and water.

Homework Equations


[itex]ΔS=\frac{Q}{T}[/itex]
[itex]Q=mcΔT[/itex]

The Attempt at a Solution


Change of entropy for the cobber block:
[itex]ΔS=S_2-S_1=\int^2_1\frac{dQ}{T}=\int^{T_2}_{T_1}mc\frac{dT}{T}=mc \ln\frac{T_2}{T_1}=3.5kg\cdot 390\frac{J}{kg K}\left ( \ln\frac{303 K}{373 K} \right )=-284 J/K[/itex]

Change of entropy for the water:
[itex]ΔS=S_2-S_1=\int^2_1\frac{dQ}{T}=\int^{T_2}_{T_1}mc\frac{dT}{T}=mc \ln\frac{T_2}{T_1}=0.8kg\cdot 4190\frac{J}{kg K}\left ( \ln\frac{303 K}{273 K} \right )=349 J/K[/itex]

Total change of entropy:
[itex]ΔS=-284 J/K+349 J/K=+65 J/K[/itex]
 
on Phys.org
Hello PhyIsOhSoHard,

PhyIsOhSoHard said:

Homework Statement


A 3.5 kg block of cobber at 100 degrees celsius (373 K) is put in 0.8 kg water at 0 degrees celsius (273 K).
The equilibrium temperature is 30 degrees celsius (303 K).
Calculate the change of entropy for the system of cobber and water.

Homework Equations


[itex]ΔS=\frac{Q}{T}[/itex]
[itex]Q=mcΔT[/itex]

The Attempt at a Solution


Change of entropy for the cobber block:
[itex]ΔS=S_2-S_1=\int^2_1\frac{dQ}{T}=\int^{T_2}_{T_1}mc\frac{dT}{T}=mc \ln\frac{T_2}{T_1}=3.5kg\cdot 390\frac{J}{kg K}\left ( \ln\frac{303 K}{373 K} \right )=-284 J/K[/itex]

Change of entropy for the water:
[itex]ΔS=S_2-S_1=\int^2_1\frac{dQ}{T}=\int^{T_2}_{T_1}mc\frac{dT}{T}=mc \ln\frac{T_2}{T_1}=0.8kg\cdot 4190\frac{J}{kg K}\left ( \ln\frac{303 K}{273 K} \right )=349 J/K[/itex]

Total change of entropy:
[itex]ΔS=-284 J/K+349 J/K=+65 J/K[/itex]

Your approach looks correct to me. :approve:

However, I would be careful with rounding numbers too early. For example, some of the sources I see online for the specific heat capacity for copper are around 386 J/(Kg K) rather than 390.

As a rule of thumb, try to keep a couple of extra significant digits around in the intermediate calculations, then round to the appropriate number of significant figures at the very end. :smile:
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K