Ch20; Heat and the First Law of Thermodynamics
Crystalline materials change phase -- melt and freeze or vaporize and condense -- at a single, fixed temperature.
Energy is required for a change of phase of a substance. The ratio of the energy to the mass of the substance involved is called the latent heat of the substance. This is much like the specific heat we have just discussed. It is called latent heat because there is no change or difference in temperature.
Q = m L
Latent heat of fusion Lf describes the heat necessary to melt (or freeze) a unit mass of a substance.
Latent heat of vaporization Lv describes the heat necessary to vaporize (or condense) a unit mass of a substance.
We have already seen
Q = c m T
For a change of state, we now have
Qf = m Lf
Qv = m Lv
Now, for the fun part!
Determining the Equilibrium Temperature
We are now ready to take the ideas of this section and make some interesting calculations or useful predictions. This is but another example of energy conservation.
For a thermally isolated system, heat may flow from one object to another but the total heat flow must be zero. That is important. For a system of two object, this means
Q1 + Q2 = 0
If there is no change of state, this means
Q1 = c1 m1 T1 = c1 m1 ( Tf - T1i )
Q2 = c2 m2 T2 = c2 m2 ( Tf - T2i )Mix 0.15 kg of 10oC milk into 0.6 kg of 95oC coffee. Both are essentially water with a specific heat capacity of 4186 J/(kg Co). What is the equilibrium temperature?
If there is a change of state, we must include the latent heats involved.
Q = c m T + m Lor even Q = c1 m T + m L + c2 m TThere is no single "general equation" to memorize. The best thing -- the only thing -- to do is to understand that this is just energy conservation and look at all the processes that cause heat loss and all the processes that cause heat gain. Set those two equal to each other -- or set the sum of them equal to zero.Let's make some iced tea! Stir 0.25 kg of ice, initially at - 5oC, into 0.5 kg of tea, initially at 45oC. What is the final temperature?
The specific heat of (liquid) water is cw = 4186 J/(kg Co). The heat of fusion for ice or water is Lf = 3.33 x 105 J/kg. The specific heat of (frozen) ice is ci = 2090 J/(kg Co). Notice that the specific heat is different for ice and water!
Working many examples is the best way to become comfortable with Equilibrium Temperature problems.
(c) Doug Davis, 2002; all rights reserved
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