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Monday, April 6, 2009

Cooling Off

So technically, I'm not supposed to be today's scribe, but its not like anyone else took scribe notes but it doesn't really matter since today's class was pretty straight forward. And for the record, the Scribe List is always up to date 8).

Now, it's not really cooling off, since it's spring! Well, with all the left over snow, it doesn't really look like spring break, but you get what I mean. Anyways, we started the class talking about how we're in the home stretch (That's right guys, there's only less than a month left before exams) and we need to start getting some practice in. There's about 72 exam questions we can work on (found over on the side here --->) and ideally, we should be getting in about 3 questions a day. Remember to not erase any of your work, that way, when you consult with Mr. K or other students, we can analyze the work to see where you went wrong (if you were wrong at all in the first place).

Anyways, let's get to the main stuff..

We started by finishing off last week's question about the roast. We found that k = 1/2ln(5/12)

So now that we know A and k, we can now solve for any time where t = time. So we then had to find how long would it take for the temperature of the roast to fall to 21 degrees Fahrenheit.

In place of T we put 21, because that's the temperature we want to find the time for, subtract it from 20. That leaves us with one. We move the 48 from the other side to isolate e^1/2ln(5/12). We then took the ln of both sides and solved for t and got approximately 8.844 hours.

So next on slide 7, we started cooling soda's outside. So now we want to find how long would it take for the drinks to cool to 35 Degrees Fahrenheit. So now, we have some initial values that we can work with.

The temperature outside is 25 Degrees Fahrenheit(we checked the thermometer outside), and the temperature inside is 72 Degrees Fahrenheit. We put one of the cans outside for 30 minutes or 1/2(0.5) an hour and measured it's temperature to be 60 Degrees Fahrenheit.

So, T(30mins) = 60 Fahrenheit (or T(0.5hours) = 60 Fahrenheit), and T(0) = 72 Fahrenheit. Because the drink is cooling at a constant rate, we can make a function. dT/dt, where T = Temperature and t = Time.
dT/dt = k(T-25)

We antidifferentiated the function and let e^c = A. ( You will recall this procedure from the last scribe post).
So we want to know what k is. So we use T(0) = 72. The e^(k(0)) will become 1 and we're left with A. So 72 - 25 leaves us with 47. Now we have our A value. To find the k value, we now use (in this case) T(30) = 60.
So we get 60 - 25 = 47e^(k(30)). We then isolated e^30k by dividing 47 from both sides. We now have 35/47 = e^30k. Finding k, we find the ln of both sides and now have 1/30ln(35/47) = k. Using the calculator, we find that k is approximately -0.0098.

So now that we have k and A, we can now find the time it takes for the drinks to cool to 35 Degrees Fahrenheit. So we set up the equation above. The first two lines are pretty straight forward, so I'll just tell you what's happening in the last three.
After plugging and chugging, we get 10/47 = e^-o.0098t and take the ln of both sides to get ln(10/47) = -0.0098t. Divide 0.0098 from both sides and you get t is approximately 157.486. Now that's not a pretty number to say unless you're talking movies, so, we convert it to be 2 hours, 32 minutes and 30 seconds. Voila!

Homework is Exercise 9.4.

Next scribe is Paul.

Now I gotta jet. Bye.

~Rence - Out

1 comment:

Lani said...

Hi there,

Now that you're on the homestretch, which IMHO is absolutely no time for cooling off, what plans and strategies are you using to assure that you can achieve to your highest potential? other than practicing?