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 $T_{f}$. Calculate $T_{f}$.

Initial State: Different temperatures, not in contact

Final State: Blocks have come to thermal equilibrium, same $T_{f}$

300 gram block of aluminum which is at a temperature of 500 K

650 gram block of iron which is at a temperature of 350 K

Aluminum is placed on iron

Both in insulated enclosure

Specific heat capacity of aluminum is approximately 1.0 J/K/gram

Specific heat capacity of iron is approximately 0.42 J/K/gram

Metal blocks reach same temperature

$T_{f}$ common temperature

Assume no other energy transfers.

System: The two blocks

Surroundings: Due to the insulation, no objects exchange energy with the chosen system

Energy Principle: ${\Delta E_{Al}} + {\Delta E_{Fe}} = 0$

${\Delta E_{sys}} = W$

$mC{\Delta T} = W$

Due to the energy principle

${\Delta E_{Al}} + {\Delta E_{Fe}} = 0$

The total energy of the two blocks does not change, because there is no energy transfer from or to the surroundings.

Therefore we want to solve for $T_{f}$ which is the final temperature for both blocks.

$m_{1}C_{1}(T_{f} - T_{1i}) + m_{2}C_{2}(T_{f} - T_{2i}) = 0$

Substitute in values for all known variables and solve for $T_{f}$.

$(300g)(1.0 J/K/g)(T_{f} - 500) + (650 g)(0.42 J/K/g)(T_{f} - 350) = 0$

Solving for the final temperature, we find $T_{f} = 429 K$.

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.