Thermal Equilibrium of a system

1. The problem statement, all variables and given/known data
A combination of .25kg of H2O at 20C, .4kg of Al at 26C and .1kg of Cu at 100C is mixed in an insulated container and allowed to come to thermal equilibrium. Ignore any energy transfer to or from the container and determine the final T of the mixture.


2. Relevant equations



3. The attempt at a solution

At first this seemed really simple to me I just figured out if the T would be below or above 26 to determine which parts will lose energy and which will gain.
So it turns out that Tf will be less than 26 because the energy required to heat up water to 26 is greater than the copper can supply.

Setting up the equation is where I get lost.

If the energy lost by the Al and Cu must equal the energy gained by the water then shouldn't the equation be:

Q(Cu)+Q(Al)=Q(H2O)

where Q= mCΔT

The book has it set up as

Q(Cu)+Q(Al)+Q(H2O)=0

which is really confusing me.

Ok so if I let the lost energy be negative I get the same thing as the book does but then it seems to me like I'd get the same answer in every case.

The book's way:
-Q(Cu)-Q(Al)=Q(H2O)
Q(Cu)+Q(Al)+Q(H2O)=0

But then let's say Tf > 26 then we get

-Q(Cu)=Q(Al)+Q(H2O)

Which simplifies to

Q(Cu)+Q(Al)+Q(H2O)=0

exactly the same thing! That doesn't make any sense to me.
 

BvU

Science Advisor
Homework Helper
11,703
2,537
One material's loss is another material's gain. And a negative gain is a loss. You have a tendency to want to use positive numbers only, but the book allows negative Q too. Which is good if you consider that a ΔT can be < 0 !
 
18,629
3,597
Let Q(Cu) be the energy gained by copper.
Let Q(Al) be the energy gained by aluminum.
Let Q(H2O) be the energy gained by water.
If any of these turn out to be negative, it just means that that particular species lost energy. So,
Q(Cu)=0.1CCu(T-100) is the energy gained by Cu, where T is the final temperature, even if T is less than 100.
The same kind of thing for Al and H2O in terms of T.

If you do it this way, then Q(Cu)+Q(Al)+Q(H2O)=0 makes sense, since the sum of the three energy "gains" is zero.
 
Ah that makes sense. Thx.
 

Want to reply to this thread?

"Thermal Equilibrium of a system" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Top Threads

Top