Total change in entropy, what am I doing wrong?

In summary, the cup reached equilibrium with a final temperature of 51°C. The entropy of the cup increased by 0.296567 J/K.
  • #1
asdf188
4
0
120 g insulated aluminum cup at 13°C is filled with 140 g of water at 58°C. After a few minutes, equilibrium is reached.

I found the final temperature to be 51°C.

But I am having trouble finding the total change in entropy.

I was using

13°C = 286.15°K
51°C = 324.15°K
58°C = 331.15°K

ΔS = ΔS_cup + ΔS_water
= Q_cup/T_avg + Q_water/T_avg
= (MCΔT)_cup/T_avg + (MCΔT)_water/T_avg
= (120g * 0.22 * (324.15°K - 286.15°K))/((324.15°K + 286.15°K)/2)
+ (140g * 1 * (324.15°K - 331.15°K))/((324.15°K + 331.15°K)/2)
=0.296567

But I am told this is the wrong answer. What am I doing wrong here?
 
Physics news on Phys.org
  • #2
asdf188 said:
120 g insulated aluminum cup at 13°C is filled with 140 g of water at 58°C. After a few minutes, equilibrium is reached.

I found the final temperature to be 51°C.

This is good.

I was using

13°C = 286.15°K
51°C = 324.15°K
58°C = 331.15°K

ΔS = ΔS_cup + ΔS_water
= Q_cup/T_avg + Q_water/T_avg

This last line is not good.

Have you come across this equation:

[tex] \Delta S = \int C \frac{dT}{T} = C ln \frac {T_2}{T_1} ? [/tex]
 
  • #3
asdf188 said:
120 g insulated aluminum cup at 13°C is filled with 140 g of water at 58°C. After a few minutes, equilibrium is reached.

I found the final temperature to be 51°C.

This is good.

I was using

13°C = 286.15°K
51°C = 324.15°K
58°C = 331.15°K

ΔS = ΔS_cup + ΔS_water
= Q_cup/T_avg + Q_water/T_avg

This last line is not good.

Have you come across this equation in your reading:

[tex] \Delta S = \int C \frac{dT}{T} = C ln \frac {T_f}{T_i} [/tex]

C is the specific heat of your liquid or solid. [tex] T_f [/tex] is the final temp, [tex] T_i [/tex] is the initial temp. Your use of average temperature is the mistake.
 
  • #4
Using your formula I have 691.796 J/K from 900ln(51/13) + 4186ln(51/58) but I am still wrong in my answer. Am I supposed to convert to Kelvins? or use kcal/kg?
 
  • #5
The "C" in front of the ln is the specific heat, so for the aluminum can that will be .215 cal/gm-C, (which is the same as cal/gm-K) and for the water 1 cal/gm-C. I'd use Kelvin. By my figuring you should get a rather small number.
 
  • #6
The answer is 0.94 J/K. But I have no idea how he got there.

the C for alum from our book is .22 so I got .22ln(324K/286K) + 1ln(324K/331K) = .006071 J/K

I am still missing something aren't I?
 
  • #7
asdf188 said:
The answer is 0.94 J/K. But I have no idea how he got there.

the C for alum from our book is .22 so I got .22ln(324K/286K) + 1ln(324K/331K) = .006071 J/K

I am still missing something aren't I?
Watch the signs. The dQ for the cup is positive because heat flows into the aluminum. But the dQ for the water is negative because heat leaves the water.

AM
 
  • #8
Also note that C is cal/gm-K. while from its definition, entropy has units of cal/K . That should be a hint to you that there's one more operation for you to do. Multiply each term by the appropriate mass and your units will come out correctly.
 

Related to Total change in entropy, what am I doing wrong?

1) What is entropy and why is it important in science?

Entropy is a measure of the disorder or randomness in a system. In science, it is important because it helps us understand the direction and magnitude of changes in a system, and is a fundamental concept in thermodynamics and statistical mechanics.

2) How is total change in entropy calculated?

Total change in entropy is calculated by adding the entropy changes of all components involved in a process. This can be done using the equation ΔS = ΣS_final - ΣS_initial, where ΔS is the total change in entropy, S_final is the final entropy of the system, and S_initial is the initial entropy of the system.

3) What factors affect the total change in entropy?

The total change in entropy is affected by the number of particles involved, the temperature, and the phase (solid, liquid, or gas) of the system. It is also influenced by any energy transfers or chemical reactions that occur during the process.

4) Why is it important to consider the total change in entropy in a process?

The total change in entropy is important because it determines whether a process is spontaneous or not. A spontaneous process is one that occurs without the need for external energy input, and the total change in entropy can tell us whether a process is moving towards a more disordered state (positive change in entropy) or a more ordered state (negative change in entropy).

5) What could I be doing wrong if I am getting a negative total change in entropy?

If you are getting a negative total change in entropy, it is likely that you have made a mistake in your calculations or assumptions. Double check your equations and make sure you are taking into account all components and energy transfers in the system. It is also possible that the process you are analyzing is not spontaneous and requires external energy input.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
3K
  • Introductory Physics Homework Help
Replies
11
Views
5K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
4K
  • Introductory Physics Homework Help
Replies
8
Views
5K
  • Introductory Physics Homework Help
Replies
2
Views
4K
Back
Top