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 calledlatentheat because there is no change or difference in temperature.Q = m L

Latent heat of fusion Ldescribes the heat necessary to melt (or freeze) a unit mass of a substance._{f}

Latent heat of vaporization Ldescribes the heat necessary to vaporize (or condense) a unit mass of a substance._{v}## More Graphs

We have already seenQ = c m T

For a change of state, we now haveQ_{f}= m L_{f}

andQ_{v}= m L_{v}

Now, for thefun part!

Determining the Equilibrium TemperatureWe are now ready to take the

ideasof this section and make someinteresting calculationsoruseful predictions.This is but another example ofenergy conservation.For a

thermally isolated system,heat may flow from one object to another but thetotal heatflow must be zero. That is important. For a system of two object, this meansQ _{1}+ Q_{2}= 0If there is no change of state, this means

Q _{1}= c_{1}m_{1}T_{1}= c_{1}m_{1}( T_{f}- T_{1i})Q

_{2}= c_{2}m_{2}T_{2}= c_{2}m_{2}( T_{f}- T_{2i})Mix 0.15 kg of 10^{o}C milk into 0.6 kg of 95^{o}C coffee. Both are essentially water with a specific heat capacity of 4186 J/(kg C^{o}). What is theequilibrium temperature?

If there is a change of state, we must include the latent heats involved.Q = c m T + m L or evenQ = c There is no single "general equation" to memorize. The best thing -- the_{1}m T + m L + c_{2}m Tonlything -- to do is tounderstandthat this is justenergy conservationand look at all the processes that cause heatlossand all the processes that cause heatgain. Set those two equal to each other -- or set the sum of them equal tozero.Let's make some iced tea!Stir 0.25 kg of ice, initially at - 5^{o}C, into 0.5 kg of tea, initially at 45^{o}C. 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 10^{5}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

manyexamples is the best way to become comfortable with Equilibrium Temperature problems.(c) Doug Davis, 2002; all rights reserved

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