When a block of cold carbon
steel touches a hot aluminum
block in an insulated
chamber, both blocks eventually reach thermal
equilibrium at a temperature between
their initial temperatures. |
|
Determine the equilibrium temperature if the aluminum block has a mass of 10 kg
and an initial temperature of 380oC and the carbon
steel block has a mass of 25 kg
and an initial temperature of 10oC. |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Read : |
The mass and initial temperature of each
block are given. We know that the equilibrium temperature of the blocks must lie between the two intial temperatures. We will need to lookup
the heat capacity or specific heat of both steel and aluminum. Then, we can apply the 1st Law to a system made up of the two blocks. No work or heat
transfer crosses the boundary of this system during the approach to equilibrium, so the only unknown in the equation is the final, equilibrium temperature. So, we can solve for
it and evaluate it. |
|
|
|
|
|
|
|
|
|
|
|
|
Diagram: |
A diagram is optional
here. The system and process are fairly simple. |
|
|
Given: |
msteel |
25 |
kg |
|
|
|
mAl |
10 |
kg |
|
|
Tsteel,1 |
10 |
oC |
|
|
|
TAl,1 |
380 |
oC |
|
|
|
|
|
|
|
Find: |
T2 |
??? |
oC |
|
|
|
|
|
|
|
|
|
|
|
|
Assumptions: |
1 - |
Steel and aluminum have constant heat capacities. |
|
|
|
2 - |
No heat is exchanged with the surroundings by either
the steel or the aluminum. |
|
|
|
3 - |
Steel and aluminum are both
incompressible, so
this process is a constant volume
process. |
|
|
|
|
|
|
|
|
|
|
|
|
Equations
/ Data / Solve: |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
We begin by writing
the 1st Law and we choose as
our system the steel and the aluminum. |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Eqn 1 |
|
|
|
|
|
|
|
|
|
|
|
|
|
By cleverly selecting
our system, Q = 0 and W = 0. This makes the solution simpler. |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Because both steel and aluminum are assumed to be incompressible with constant heat capacities: |
|
|
Eqn 2 |
|
|
|
|
|
|
|
|
|
|
|
|
|
Substitute Eqn 2 into Eqn
1 twice, once for steel and once for aluminum to get : |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Eqn 3 |
|
|
|
|
|
|
|
|
|
|
|
|
Notice that there is
only one T2 because in the final, equilibrium state,
the steel and the aluminum are both at the same temperature ! |
|
|
|
|
|
|
|
|
|
|
|
|
|
Now, solve Eqn 3 for T2: |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Eqn 4 |
|
|
|
|
|
|
|
|
|
|
|
|
|
Now, solve Eqn 4 for T2. |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Eqn 5 |
|
|
|
|
|
|
|
|
|
|
|
|
|
Before we can evaluate
T2, we must lookup the specific heat of steel
and of aluminum. |
|
|
|
|
|
|
|
|
|
|
|
For carbon steel, I found : |
|
|
|
CP,st |
0.49 |
kJ/kg-K |
|
|
For incompressible solids, CV = CP, so : |
|
CV,st |
0.49 |
kJ/kg-K |
|
|
|
|
|
|
|
|
|
|
|
For aluminum, I found : |
|
|
|
CP,Al |
0.91 |
kJ/kg-K |
|
|
|
|
|
|
CV,Al |
0.91 |
kJ/kg-K |
|
|
|
|
|
|
|
|
|
|
|
Now, we can finally
plug numbers into Eqn 5 and
evaluate T2: |
T2 |
167.7 |
oC |
|
|
|
|
|
|
|
|
|
|
Verify: |
None of the assumptions
can be verified. |
|
|
|
|
|
|
|
|
|
|
|
|
Answers : |
T2 |
168 |
oC |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|