What is the total entropy change of the system tea + air?

In summary: SolveIn summary, the conversation discusses the process of making tea with 0.250 kg of 89.0 °C water and letting it cool to room temperature before drinking it. The entropy change of the water while cooling is calculated using the equation m*c*ln(T2/T1) = -221 J/K. The cooling process is assumed to be isothermal for the air in the kitchen, and the change in entropy of the air is calculated using the equation deltaS = Q/T. The total entropy change of the system tea + air can then be found by adding the entropy changes of the water and air. The solution is found by using the equation deltaS = Q/T with Q being the heat flow from the
  • #1
danni7070
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Homework Statement


You make tea with 0.250 kg of 89.0 °C water and let it cool to room temperature (20.0 °C) before drinking it.

1)Calculate the entropy change of the water while it cools.
2)The cooling process is essentially isothermal for the air in your kitchen. Calculate the change in entropy of the air while the tea cools, assuming that all the heat lost by the water goes into the air.
3)What is the total entropy change of the system tea + air?

Homework Equations



m = 0.25kg
T1 = 362K
T2 = 293K



The Attempt at a Solution



1) I did m*c*ln(T2/T1) = -221 J/K And it's correct

2) Now I'm not sure what to do here.

Isothermal means that T = constant so is it 293K ?

And what eq. am I suppose to use and why isn't the volume of the room a factor here?

What I tried to do was to find the heat Q = m*L_f (L_f = 3.34*10^5 J/kg)

So Q = 0.25*L_f = 83500 J

Then (delta)S = Q/T = 83500J / 293 K = 289 J/K But that's not correct and I guess the entropy should me a small number but I'm not really sure since I'm not that good at understanding "High-Tech" english!


3) If I can do 2) then 3 should not be a problem.

Thank you.

EDIT: I found out that I CAN use the eq. S = Q/T IF the process is isothermal so what am I doing wrong?
ps. I've only got one attempt left and this is the last question of the year! (Apart from the test)


EDIT2: Solution

It was correct using (delta)S = Q/T using Q = 83500 J/K

The T has to be the difference of temperature ((89°C-20°C)+273) = 342 K

So Q/T = 244 J ! And it was correct.
 
Last edited:
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  • #2
The flow of heat from the tea is [itex]\Delta Q = mc\Delta T_{tea}[/itex]. This has to be the amount of heat flowing into the air. So you don't have to know the heat capacity of the air.

So the change in entropy of the air is just:

[tex]\Delta S_{air} = \Delta Q /T_{air} = mc\Delta T_{tea}/T_{air}[/tex]

AM
 
  • #3


I'm sorry for wasting your time. I'll try to think harder before I post a thread in the future!

No problem, it's always good to ask for clarification if you're unsure about something. Here's a response to your questions:

2) Yes, in this case, T = 293K because the air is kept at a constant temperature during the process. The volume of the room is not a factor because we are only concerned with the change in entropy of the air, not the total entropy of the room.

To calculate the change in entropy of the air, we can use the formula S = Q/T, where Q is the heat absorbed by the air and T is the temperature at which it is absorbed. In this case, the heat absorbed by the air is equal to the heat lost by the water (since we assume all the heat lost by the water goes into the air) and the temperature at which the heat is absorbed is 293K. So the change in entropy of the air is (83500 J)/(293 K) = 285 J/K.

3) The total entropy change of the system is simply the sum of the entropy changes of the water and the air. So in this case, the total entropy change is -221 J/K + 285 J/K = 64 J/K.

I hope this helps! Let me know if you have any other questions.
 

1. What is entropy and why is it important?

Entropy is a measure of the disorder or randomness in a system. It is important because it helps us understand the direction and extent of spontaneous processes, and is a fundamental concept in thermodynamics and statistical mechanics.

2. How is entropy related to tea and air?

In the context of thermodynamics, the entropy of a system is directly related to the amount of heat energy transferred to or from the system. In the case of tea and air, the total entropy change of the system depends on the heat energy exchanged between the tea and the surrounding air.

3. What factors affect the total entropy change of the system?

The total entropy change of the system is affected by the temperature difference between the tea and air, the amount of heat energy exchanged, and the heat capacity of the substances involved. Additionally, any changes in the composition or pressure of the system can also affect the entropy change.

4. Is the total entropy change of the system always positive?

No, the total entropy change of the system can be positive, negative, or zero. If heat energy is transferred from the tea to the air, the entropy change will be negative. If heat energy is transferred from the air to the tea, the entropy change will be positive. If there is no net heat transfer, the entropy change will be zero.

5. How can we calculate the total entropy change of the system?

The total entropy change of the system can be calculated using the formula ΔS = ΔQ/T, where ΔS is the change in entropy, ΔQ is the heat energy transferred, and T is the temperature. This calculation can be applied to specific situations, such as heating or cooling the tea, to determine the total entropy change of the system.

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