# 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?

##### 1 Answer

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 your problem:

Specific heat capacity for water is

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.

(

(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 =

The temperature at which thermal equilibrium will occur is

Resource: http://www.buzzle.com/articles/thermal-equilibrium.html