# A ball of aluminum (cp = 0.897 J/g°C) has a mass of 100 grams and is initially at a temperature of 150°C. This ball is quickly inserted into an insulated cup with 100 ml of water at a temperature of 15.0°C. What will be the final, equilibrium temperature of the ball and the water?

Oct 1, 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

Specific heat capacity for water is $\text{4.18 J/g"^"o""C}$.
Heat lost by aluminum = Heat gained by water
The density of water is 1g/mL, so 100mL of water has a mass of 100g.

Q aluminum = - Q water
m x cp x (Tf - Ti) = -m x cp x (Tf - Ti)

Plug in the values pertaining to aluminum on the left side of the equation, and those pertaining to water on the right side.

($\text{100 g}$)($\text{0.897J/g°""C}$)($\text{T"_"f}$ - 150°"C") = - ($\text{100g}$)($\text{4.18J/g"°"C}$)(${\text{T}}_{f}$ - 15.0°"C")

(89.7)( Tf - 150) = - 418( Tf - 15.0) (dropped units to make the algebra easier)

89.7Tf - 13455 = - 418Tf + 6270 (negative x negative = positive)

Combine like terms by adding 418Tf to both sides, and adding 13455 to both sides.

507.7Tf = 19725

Tf = $\text{38.9"^"o""C}$ (divided 19725 by 507.7; put the remaining units back at the end)

The temperature at which thermal equilibrium will occur is $\text{38.9"^"o""C}$.