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?
Answers
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
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.
(
100 g
)(
0.897J/g°
C
)(
T
f
-
150
°
C
) = - (
100g
)(
4.18J/g
°
C
)(
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 =
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
38.9
o
C
.