# A 12 gram piece of aluminum (cp = .215 cal/g°C) at 70°C is placed in a beaker that contains 35 grams of water at 15°C. At what temperature will they come to thermal equilibrium?

Oct 2, 2014

When two substances having different temperatures are introduced or kept together, heat energy, Q, flows from a substance at higher temperature to a substance at lower temperature. Also, heat continues to be transferred until their temperatures are equalized, at which point the substances are in thermal equilibrium. In a closed system, the amount of energy lost is equal to but opposite the amount of energy gained.

Thermal equilibrium formula

Q = m x cp x ∆T or Q = m x cp x (Tf - Ti)

where Q = Heat Flow (Heat lost or Heat gained)
m = Mass of the substance
cp = Specific heat capacity
Tf = Final temperature
Ti = Initial temperature
∆T = (Tf - Ti) = Difference in temperature

For this problem:

Specific heat capacity for water = $\text{1.00 cal/g"^"o""C}$
Q aluminum = -Q water
m x cp x (Tf - Ti) = -m x cp x (Tf - Ti)

Plug in the values for aluminum on the left side and water on the right.

(12g)(0.215)(Tf - 70) = -(35g)(1.00)(Tf - 15) (dropped units to make the algebra easier)

(2.58)(Tf - 70) = -(35)(Tf - 15)

2.58Tf - 180.6 = -35Tf + 525) (distributive property of multiplication and a negative times a negative gives a positive)

Combine terms by adding 180.6 and 35Tf to both sides.
37.58Tf = 705.6

Divide both sides by 37.58.
Tf = $\text{18.8"^"o""C}$ (put remaining units back)

The temperature at which thermal equilibrium will occur is $\text{18.8"^"o""C}$