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.

http://ift.tt/1kiiUhA

Leave a comment

Your email address will not be published.


*


Show Buttons
Hide Buttons