Thermal Equilibrium of a system

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Homework Help Overview

The problem involves determining the final temperature of a mixture consisting of water, aluminum, and copper in an insulated container, focusing on thermal equilibrium. The specific heat capacities and mass of each material are provided, and the initial temperatures vary significantly.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of energy equations for the system, questioning the signs of energy gains and losses. There is confusion regarding the formulation of the energy balance equation and how to interpret the signs of the quantities involved.

Discussion Status

Some participants have provided insights into the interpretation of energy gains and losses, suggesting that negative values can represent energy loss. There is an ongoing exploration of how to correctly apply the energy balance equation, with different interpretations being discussed.

Contextual Notes

Participants are navigating the implications of the signs in the energy equations and how they relate to the final temperature, with some uncertainty about the correct approach to setting up the problem.

Feodalherren
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Homework Statement


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.


Homework Equations





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.
 
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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 !
 
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.
 
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Ah that makes sense. Thx.
 

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