G with which do we start well without any further

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Unformatted text preview: Example 4: In the last few problems, the starting point has always been the number with one single unit (as opposed to this per that), so here's a problem that is slightly different. The velocity of light (as we've seen) is 3.0x1010 cm / sec. What is this speed in miles / hour? GIVEN: 3.0x1010 cm/sec FIND: velocity in miles / hour CONVERSION FACTORS: We'll need 2.54 cm / in, 12 in / ft, 5280 ft / mile, 60 sec / min and 60 min / hour OK, in this case the starting point is pretty obvious, but this won't always be the case. If you're not sure where to start, start anywhere and just work it out. This can lead to the problem of having the correct units but in the inverse arrangement. This problem can be solved by simply inverting the answer, as will be seen in the following example. As for the present example, let's work it out. 3.0x1010 cm inch foot mile 60 sec 60 min * * * * * = 6.7x108 mile hour sec 2.54 cm 12inch 5,280 foot min hour Notice in this one that we let the unit "sec" slide along until we first converted from cm to mile. Example 5: Finally, not all problems will go along seamlessly. One problem that shows up periodically is when you get the correct units, but in the inverse order (that is the numerator and denominator are switched). There are two ways to solve this difficulty, one is to go back and start over again with the starting point put in the inverse way that you started before (for instance, if you started with feet / sec, then we would use sec / feet). The other is to notice that taking 1. / the answer will put the units in the right way, which is simply one divided by the answer. You may want to initially try both ways, one to verify the other, until you are comfortable with this little "factor label trick". Here's an example problem. If we add 150,000 cal to 100. g of water with a specific heat of 1.0 cal / g o C, what will be the temperature change of the water in degrees Celsius? GIVEN: 150,000 cal, 100. g, 1.0 cal / g 0 C FIND: Temperature change in o C Dakota State University page 223 of 232 Factor Label Method General Chemistry I and II Lab Manual CONVERSION FACTORS: no additional conversion factors needed Notice that we have not one, but two single unit numbers (150,000 cal and 100. g). With which do we start? Well, without any further information, it's not clear, so we'll start with one of them, say 100 g, and we'll just keep going. So, let's set it up. 100.g * 10cal . 1 * = 6.67x10−4 oC− 1 o g C 150,000cal There are a couple of things to note about this problem. The first is that the last term is written as 1/150,000 cal. I knew I needed calories in the denominator to cancel calories in the numerator, but 150,000 calories is a single unit number. Thus, I recognize that 150,000 cal is equivalent to 150,000 cal / 1, so to get calories in the numerator, I just "flip" this number around. I did NOT add any superfluous or artificial units to the "1" in the numerator. When we write 150,0...
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This note was uploaded on 09/18/2012 for the course CHEMISTRY 1010 taught by Professor Kumar during the Fall '11 term at WPI.

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