Change in entropy

  • #1
Pouyan
103
8
1 kg of iron with the temperature of 323 K to 293 K is cooled by dipping into a large water bath temperature of 293 K. What is the total entropy change?


Relevant equations
dS = mcdT/T
S=mc *ln(T2/T1)
C iron = 450J/(kgK)
C water= 4200 J/(kgK)
m water = I see in my book 1 kg in solution

The attempt at a solution
I see in my book that they solved this problem like this :

ΔS for Iron = mc ln(T2/T1)= 1*450* ln(293/323) (J/K)= -43.9 J/K


I don't have problem to understand this but I see for water that they did like this :

ΔS for water = mc*(T2-T1)/T1 = 1*450*30/(293) = 46.1 J/K

Here they used C iron to calculate ΔS for water!

And the total increase of entropy in this system is
46.1-43.9 = 2.2J/K

I have two questions

1)why can not we use integrals to find the entropy change in the water as we did for the iron ?

2)and the second question is that why we use the specific heat capacity of iron when we'd calculate the entropy change in the water?!
 

Answers and Replies

  • #2
Pouyan
103
8
Can someone help me please with this question ?!
 
  • #3
22,219
5,120
Can someone help me please with this question ?!
Yes. I can answer all your questions, but not right now. Be back later.
 
  • #4
22,219
5,120
In this problem, they assume that the water bath is an infinite ocean at 293 K. In thermo, that's what they mean when they refer to a water bath or a reservoir (or an infinite reservoir). As such, the product of the mass of the reservoir times the heat capacity of the reservoir is considered infinite. As such, when heat is added to the reservoir, its temperature doesn't change.

We can determine the change in entropy of the water bath in two equivalent ways. Both methods give the same result. Let Q represent the amount of heat removed from the iron that goes into the reservoir. This is equal to mironcironΔTiron. This is exactly the same as the amount of heat entering the water bath.

Method 1:
Since the temperature of the water bath doesn't change and its entropy change is the integral of dQ/Tbath, the entropy change of the reservoir is ΔS=Q/Tbath

Method 2: Here we assume that the reservoir has a finite size (i.e., product of mass times heat capacity) rather than infinite size. Then we analyze what happens in the limit if we allow the mass times heat capacity to become infinite. Let Tinit represent the temperature of the reservoir before it equilibrated with the chunk of iron. If the final temperature of the bath was Tbath, the amount of heat added to the bath was Q, and the product of mass and heat capacity of the bath was (mc)bath, what was initial temperature of the bath Tinit? Using your equation that involved integrals, what was the change in entropy for this case of finite reservoir size?

After you provide these answers, we'll continue.

Chet
 
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  • #5
Pouyan
103
8
Thanks!

the entropy for water is : ΔS for water = mc*(T2-T1)/T1 = 1*450*(323-293)/(293) = 46.1 J/K
they used (mc) iron ,

The initial temperature of water bath is 293 K.
 
  • #6
22,219
5,120
Thanks!

the entropy for water is : ΔS for water = mc*(T2-T1)/T1 = 1*450*(323-293)/(293) = 46.1 J/K
they used (mc) iron ,

The initial temperature of water bath is 293 K.
No. For the case described in Method 2 (in which the water bath is finite size rather than infinite size), the initial temperature of the water bath had to be:

$$T_{init}=293-\frac{Q}{(mc)_{bath}}$$
where $$Q=450(323-293)=(mc)_{bath}(293-T_{init})$$

So, what does your equation for ΔS predict for this case (i.e., ##ΔS=(mc)_{bath}\ln(293/T_{init})##)? That is, make the substitution for Tinit. What do you get?

Chet
 
  • #7
Pouyan
103
8
Sorry but it's not in the exercise in my book. The temperature of water is 293 K at start
 
  • #8
22,219
5,120
Sorry but it's not in the exercise in my book. The temperature of water is 293 K at start
In method 2, I've changed the problem a little to help you understand what's happening. You said that you wanted to understand, correct? If you're no longer intetested, OK. Have a nice day.:smile:
 
Last edited:
  • #9
22,219
5,120
I've decided that, just for the sake of completeness, I am going to complete the analysis of Method 2. So, here goes.

Starting from post #6,
$$ΔS=(mc)_{bath}\ln\left(\frac{293}{T_{init}}\right)=(mc)_{bath}\ln\left(\frac{293}{293-\frac{Q}{(mc)_{bath}}}\right)$$
This can be rewritten as:$$ΔS=-(mc)_{bath}\ln\left(1-\frac{Q}{293(mc)_{bath}}\right)$$
Now, if we take the limit of this expression as the thermal inertial of the reservoir becomes infinite (i.e., ##(mc)_{bath}→∞##), we obtain:
$$ΔS=\frac{Q}{293}$$

This completes then analysis of Method 2.

Chet
 
  • #10
CrazyNinja
173
60
@Chestermiller ... Hello!

I was just looking this problem up to understand the concept of entropy better and I was wondering if you could tell me what the advantages of the second method are over the first. Is it only that it allows us to let us use the limit (as in let it tend to anything we need it to?)
 
  • #11
22,219
5,120
Method 2 was worked out so that we could show how the results for an arbitrary reservoir size could be extended to the limit of an infinite reservoir. For an arbitrary object (or reservoir) experiencing only a change in temperature, we can derive the entropy change by devising a reversible process to bring the object from its initial temperature to its final temperature. One way of doing this is to contact the object with an infinite sequence of reservoirs of slightly different temperatures, running from the initial temperature to the final temperature. This leads to the equation $$\Delta S=mC\ln(T_f/T_i)$$
 
  • #12
Hangtime
11
0
Can a simplistic understanding of entropy also be stated as the natural expenditure of a process falling to a lower energy state?
 

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