Change of entropy of a system

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fluidistic
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Homework Statement


A kilogram of water at 0°C is put in contact with a reservoir at 100°C. When the water reaches 100°C, what is the change of entropy of the water, the reservoir, and the whole system?


Homework Equations


[tex]\Delta S = \int \frac{dQ}{T}[/tex].


The Attempt at a Solution



I realize that the change of entropy of the water is positive while the change of entropy of the tank must be negative. However the sum of the changes of entropy must be positive since it's an irreversible process.
However I don't reach this. What I did was, for the water : [tex]\Delta S =mc \int_{T_0}^{T_{100}} \frac{dT}{T}=1000g \cdot \frac{1 cal}{g K}\ln \left ( \frac{373}{273} \right )=74.66J/K[/tex].

Now the change of entropy of the tank : [tex]\Delta S = \frac{Q}{T}[/tex] since [tex]T[/tex] is constant.
[tex]Q=-mc100K \Rightarrow \Delta S=-1120.64J/K[/tex].

Sidenote : I considered 1cal=4.18 J.

Where did I go wrong?
 

Answers and Replies

  • #2
rl.bhat
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In isothermal condition ΔS = 0.
There is no change in the temperature of the reservoir. So ΔS = ...?
 
  • #3
rl.bhat
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In isothermal condition ΔS = 0.
There is no change in the temperature of the reservoir. So ΔS = ...?
 
  • #4
fluidistic
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In isothermal condition ΔS = 0.
There is no change in the temperature of the reservoir. So ΔS = ...?
The change of entropy of the reservoir would be 0 but ... are you sure? I've the following solved problem : A reservoir at 300°C is put in contact with a reservoir at 0°C. 20kJ of heat flows irreversibly from the first to the second. Calculate the variation of entropy of the Universe.

Solution : Change of entropy of the first reservoir : [tex]-\frac{20kJ}{573K}=-0.03490kJ/K[/tex].
The for the second reservoir, [tex]\Delta S=0.07322kJ/K[/tex].
Adding them up, [tex]\Delta S _{\text{total}}=38.3J/K[/tex] which is greater than 0.
It's the solution of a test that 5 professors gave.
The temperature of the reservoirs doesn't change, they are assumed to be infinite, although it is not said in the question. (That's why I couldn't solve it!)
 
  • #5
kuruman
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[tex]\Delta S =mc \int_{T_0}^{T_{100}} \frac{dT}{T}=1000g \cdot \frac{1 cal}{g K}\ln \left ( \frac{373}{273} \right )=74.66J/K[/tex].

Sidenote : I considered 1cal=4.18 J.

Where did I go wrong?

Look at your numbers. To convert from calories to Joules you divided by 4.18 instead of multiplying. Redo the calculation and you will get a number greater than 1120 J/K. All's well with the Second Law.
 
  • #6
fluidistic
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Look at your numbers. To convert from calories to Joules you divided by 4.18 instead of multiplying. Redo the calculation and you will get a number greater than 1120 J/K. All's well with the Second Law.

Ah thanks... I don't have anything else to say.
 

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