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Example: Thermal Equilibrium
A 300 gram block of aluminum at temperature 500 K is placed on a 650 gram block of iron at temperature 350 K in an insulated enclosure. At these temperatures the specific heat capacity of aluminum is approximately 1.0 J/K/gram, and the specific heat capacity of iron is approximately 0.42 J/K/gram. Within a few minutes the two metal blocks reach the same common temperature Tf. Calculate Tf.
Facts
Initial State: Different temperatures, not in contact
Final State: Blocks have come to thermal equilibrium, same Tf
Lacking
Approximations & Assumptions
Assume no other energy transfers.
Representations
System: The two blocks
Surroundings: Due to the insulation, no objects exchange energy with the chosen system
Energy Principle: ΔEAl+ΔEFe=0
ΔEsys=W
mCΔT=W
Solution
Due to the energy principle
ΔEAl+ΔEFe=0
The total energy of the two blocks does not change, because there is no energy transfer from or to the surroundings.
m1C1(Tf−T1i)+m2C2(Tf−T2i)=0
(300g)(1.0J/K/g)(Tf−500)+(650g)(0.42J/K/g)(Tf−350)=0
Solving for the final temperature, we find Tf=429K.
In words, what happens in this process is that the aluminum temperature falls from 500 K to 429 K, and the iron temperature rises from 350 K to 429 K. The thermal energy decrease in the aluminum is equal to the thermal energy increase in the iron.