Thermal Equilibrium of a system

[Feodalherren]

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.

 

[BvU]

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 !

 

[Chestermiller]

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)=0.1C_Cu(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.

 

[Feodalherren]

Ah that makes sense. Thx.

 

[HalfLostAndSearching]

I would approach this problem by first understanding the concept of thermal equilibrium and how it applies to this specific system. Thermal equilibrium is a state in which there is no further transfer of heat between two objects or systems in contact with each other. In this case, the insulated container ensures that there is no transfer of heat to or from the surroundings, allowing the system to reach thermal equilibrium.

To determine the final temperature of the mixture, we can use the principle of energy conservation. The total energy of the system before and after mixing must be the same. This means that the energy lost by the Al and Cu must equal the energy gained by the water.

Using the equation Q = mCΔT, where Q is the heat energy, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature, we can set up the following equation:

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

Where Q(Cu) is the energy lost by the copper, Q(Al) is the energy lost by the aluminum, and Q(H2O) is the energy gained by the water.

Solving for Q(H2O), we get:

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

Plugging in the values given in the problem, we get:

Q(H2O) = -0.1kg(387J/kg°C)(100°C - Tf) - 0.4kg(897J/kg°C)(26°C - Tf)

Where Tf is the final temperature of the mixture.

Solving for Tf, we get:

Tf = 42.4°C

Therefore, the final temperature of the mixture is 42.4°C. This is lower than the initial temperature of the aluminum and copper, indicating that the water gained more energy from the aluminum and copper than they lost. This makes sense as the specific heat capacity of water is higher than that of aluminum and copper, meaning it requires more energy to raise its temperature.