University of Wisconsin Green Bay

You pour 320 g (0.32 kg) of boiling tea into an insulated travel mug. How many 12 g (0.012 kg) cubes of ice do you need to add in order to cool the tea to a temperature of no more than 57 oC? Hints: Tea is essentially water and so you can use any constants that you look up for water to apply to tea as well. Don’t worry about any changes to temperature of the travel mug.

• In this problem, you are first asked about a situation in which the temperature of one substance (tea) changes because of its interaction with another substance (ice). Temperature is a measure of thermal energy, so in other words thermal energy goes from the tea to the ice. This is a Conservation of Energy problem. Because only thermal energies are involved, you do not need to look at mechanical (kinetic, potential) energies.

• Unlike mechanical energy problems, no figure is needed in this case. All energy changes are from thermal energy going from the tea to the ice. A picture will not add to understanding or organization of this problem.

• The energy chain for this problem is

Thermal energy of tea → thermal energy of ice or

-ΔQtea = ΔQice
-mteacteaΔTtea = miceLof melting + micecmelted iceΔTice

Any time you understand the motion of an object by looking at its energy, you begin with a statement of Conservation of Energy. You don’t need to track kinetic and potential energies here so it is easiest to begin with an energy chain.

• In order for the tea to cool from boiling to 57oC, it must lose 57.6 kJ of thermal energy. This is enough to melt and warm 0.10 kg of ice. (0.10 kg)/(0.012 kg/cube) = 8.3 ice cubes so you must add nine ice cubes to bring the temperature of the tea below 57o degrees.

• The energy chain for this problem is

Thermal energy of tea → thermal energy in ice

There are no changes in mechanical energy and so no need to track those terms. In equation form,

-ΔQtea = ΔQice
-mteacteaΔTtea = miceLof melting + micecmelted iceΔTice

The first term in this equation represents the amount of energy the tea lost as it cooled, the second term is the amount of energy absorbed by the ice as it melted, and the final term is the amount of energy absorbed by the melted water as it warmed to 570C. Mathematically, the solution

-(0.32 kg)(4.187 kJ/kg∙K)(330 K – 373 K) = mice[334 kJ/kg + (4.187 kJ/kg∙K)(330 K – 273 K)
57.6 kJ = mice(573 kJ/kg)
0.10 kg = mice or 8.3 ice cubes

is physically reasonable.