Physics 133
Spring Quarter 2004
30 April 2004
Email the quiz back to me at aubrecht@mps.ohio-state.edu.
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The quiz is due before class today, 30 April 2004.
Explain what happens when a hot metal object is put into cold water. Why does anything happen? Specifically address this problem if the metal is copper at 100 °C and the water is at 20 °C. Can the final temperature be determined? What if there is 5.00 kg of water and 250 g of copper? What can be said then?
Gordon’s solution will appear here
Here, there is energy (thermal energy) in the warmer object that exceeds that in the colder object. The zeroth law of thermodynamics says that this wll lead to an exchange of energy somehow, resulting in thermal equilibrium being established. This means that the objects’ temperatures are the same.
In the specific case given, the hot copper will lose thermal energy to the colder water. In order to determine the final temperature, the amount of the materials is important. This is because the amount of energy transferred will depend on how much of the material there is. A microgram at a high temperature can transfer less energy than a tonne at a temperature not very much higher than ambient. We also need to know the respective specific heats, because the mass times the specific heat times the temperature change is what determiones the amount of energy transferred.
The thermal energy lost by the hot copper is
(0.250 kg) (390 J/(kg °C)) (100 °C - Tfinal).
The thermal energy gained by the colder water is
(5.00 kg) (4186 J/(kg °C)) (Tfinal - 20 °C).
Since these must be equal, we equate them and gather terms:
(0.250 kg) (390 J/(kg °C)) (100 °C) + (5.00 kg) (4186 J/(kg °C)) (20 °C) = [(5.00 kg) (4186 J/(kg °C)) + (0.250 kg) (390 J/(kg °C))] Tfinal, or
Tfinal = [(0.250 kg) (390 J/(kg °C)) (100 °C) + (5.00 kg) (4186 J/(kg °C)) (20 °C)]/[(5.00 kg) (4186 J/(kg °C)) + (0.250 kg) (390 J/(kg °C))] = 20.4 °C.
This is obviously not a very great rise in temperature because the specific heat of copper is so low and the amount of copper is so low compared to the water.
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