Friday Feb. 19, 2010
click here to download today's notes in a more printer friendly format

A couple of Bob Dylan songs ("Like a Rolling Stone" and "Stuck in Mobile with the Memphis Blues Again") before class today.

Quiz #1 was returned in class today.  This quiz counts so you should carefully check to be sure the grading was done correctly and that the points were added up correctly.  If you didn't do as well as you thought you should have there is still plenty of time to turn things around before the end of the semester.  I would suggest you stop by sometime during office hours to discuss what I think are effective quiz study strategies.

For the next couple of weeks or so we will be learning about several different forms of energy, energy transport, and the atmospheric greenhouse effect.  It is easy to get wrapped up in all the details so class started with a little bit of an introduction.

When you add energy to an object, the object will usually warm up (conversely when you take energy from an object the object will cool).  It is relatively easy to come up with an equation that allows you to figure out what the temperature change will be.



The temperature change will first depend on how much energy was added.  This is a direct proportionality, so delta E is in the numerator of the equation (delta E and delta T are both positive when energy is added, negative when energy is taken from something)

When you add equal amounts of energy to large and small  pans of water, the small pan will heat up more quickly.  The temperature change, delta T, will depend on the mass.  A small mass will mean a large delta T, so mass should go in the denominator of the equation.  

Different materials react differently when energy is added to them.  A material with a large specific heat will warm more slowly than a material with a small specific heat.  Specific heat has the same kind of effect on delta T as mass.  Specific heat is sometimes called "thermal mass" or "thermal capacity."

Here's an important example that will show the effect of specific heat (middle of p. 45)


Equal amounts of energy (1000 calories, note that calories are units of energy) are added to equal masses (500 grams) of water and soil.  We use water and soil in the example because most of the earth's surface is either ocean or land. Water has a higher specific heat than soil, it only warms up 2o C.  The soil has a lower specific heat and warms up 10o C, 5 times more than the water (there is a factor of 5 difference in the specific heats of water and soil).

These different rates of warming of water and soil have important effects on regional climate.




Oceans moderate the climate.  Cities near a large body of water won't warm as much in the summer and won't cool as much during the winter compared to a city that is surrounded by land.

The yearly high and low monthly average temperatures are shown at two locations above.  The city on the coast has a 30o F annual range of temperature (range is the difference between the summer and winter temperatures).  The city further inland (assumed to be at the same latitude and altitude) has an annual range of 60o F.  Note that both cities have the same 60o F annual average temperature.

Here's another situation where you can take advantage of water's high specific heat to moderate "micro climate."




This weekend I'm planning on planting a bunch of these tomato plants that have been growing on the window sill in my office. 




Here are a couple of last year's plants.  It can still get cold enough at night this time of year to kill tomatoes (the brocolli and lettuce in the background can handle a light frost).  So you have to protect the tomato plants.




One way of doing that is to put a "wall of water" around each plant.  They take advantage of the high specific heat of water and won't cool as much as the air or soil would during a cold night.

Adding energy to an object will usually cause it to warm.  But there is another possibility (bottom p. 45),  the object could change phase (change from solid to liquid or gas).  Adding energy to ice might cause the ice to melt.  Adding energy to water could cause it to evaporate.  The figure below is a little more detailed version of what was drawn in class.

The equation at the bottom of the figure above allows you to calculate how much energy is required to melt ice or evaporate water or sublimate dry ice.  You multiply the mass by the latent heat, a variable that depends on the particular material that is changing phase. 
The following figure wasn't shown in class.


If you add energy to or remove energy from an object, the object will usually change temperature.  You can calculate the temperature change if you know the object's mass and its specific heat.  That's the equation we used in the example calculation earlier.

We will be using the equation next in a slightly different way in a class experiment/demonstration. We will measure the temperature change and use that to determine the amount of energy lost by an object.

A student from the class was nice enough to volunteer to perform the experiment (actually the student was offered a green card in exchange for performing the experiment).  The object of the experiment was to measure the latent heat of vaporization of liquid nitrogen.  That just means measuring the amount of energy needed to evaporate a gram of liquid nitrogen.  The students that are doing Experiment #2 are measuring the latent heat of fusion of ice, the energy needed to melt one gram of ice.  You'll find the following figure on p. 45a in the photocopied Classnotes.


(a)
Some room temperature water poured into a styrofoam cup weighed 104.4 g.  The cup itself weighed 3.4 g, so we had 101.0 g of water.

(b)
The water's temperature was 21.0 C  (room temperature).

(c)
  40.0 g of liquid nitrogen was poured into the cup of water.

It takes energy to turn liquid nitrogen into nitrogen gas.  The needed energy came from the water.  This flow of energy is shown in the middle figure above.  We assumed that because the experiment is performed in a styrofoam cup that there is no energy flowing between the water in the cup and the surounding air.

(d)
After the liquid nitrogen had evaporated we remeasured the water's temperature.  It had dropped to 1.0 C.  That is a temperature drop of 21.0 - 1.0 = 20.0 C.

Because we knew how much water we started with, its temperature drop, and water's specific heat we can calculate how much energy was taken from the water.
101.0 g x 20 C x 1 cal/g C  =  2020 calories

We then divide that number by the amount of liquid nitrogen that was evaporated.
2020 calories / 40 grams = 50.5 calories per gram

A trustworthy student in the class informed us that the known value is 48 cal/g, so our measurement was pretty darn close.