Ch20; Heat and the First Law of Thermodynamics

Latent Heat

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 L_f describes the heat necessary to melt (or freeze) a unit mass of a substance.
Latent heat of vaporization L_v describes the heat necessary to vaporize (or condense) a unit mass of a substance.

More Graphs

We have already seen

Q = c m T

For a change of state, we now have

Q_f = m L_f

and

Q_v = m L_v

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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 ( T_f - T_1i )

Q₂ = c₂ m₂ T₂ = c₂ m₂ ( T_f - T_2i )

Mix 0.15 kg of 10^oC milk into 0.6 kg of 95^oC coffee. Both are essentially water with a specific heat capacity of 4186 J/(kg C^o). What is the equilibrium temperature?

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If there is a change of state, we must include the latent heats involved. Q = c m T + m L or even Q = c₁ m T + m L + c₂ m T There 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 - 5^oC, into 0.5 kg of tea, initially at 45^oC. 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 L_f = 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.

Heat Capacity			Work and Heat
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(c) Doug Davis, 2002; all rights reserved