Math 2250 Lab 1 Page 7 of 8 6. As part of the summer job at a resturant, you lea
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Question
Math 2250 Lab 1 Page 7 of 8 6. As part of the summer job at a resturant, you learned to cook up a big pot of soup late at night, just before closing time, so that there would be plenty of soup to feed customers the next day. However, the soup was too hot to be put directly into the fridge when it was ready. (The soup had just boiled at 100 °C, and the fridge was not powerful enough to accomodate a big pot of soup if it was any warmer than 20 °C) (a) Suppose that by cooling the pot in a sink full of cold water, kept running, so that its temperature was roughly constant at 5 C. Let T(t) be the temperature of the pot at any time t, measured in minutes. Use Newton's Law of cooling to write down a differential equation for T(t). Neuton's Law of cooling states that the rate of change of the temperature of an object is proportional to the difference between the object's temperature and the surrounding temperature. (b) Now suppose that by cooling the pot you were able to bring the temperature of the soup to 60 °C in 10 minutes. Use this information to find the constant of proportionality in Newton's Law. (c) How long must you run the soup pot over water in the sink before you can put the soup in the fridge?Explanation / Answer
Part (a)
Given by Newton’s Law of cooling, the differential equation would be:
dT/dt = k(T – 5), where k = constant of proportionality, ANSWER………………..(1)
[note the surrounding temperature is given to be 50C in the first sentence]
Part (b)
Rewriting (1) as dT = kTdt – 5kdt and integrating both sides,
T = kTt – 5kt + C, where C = constant of integration…………………………………(2)
At t = 0, i.e., at the start of cooling. T = 100 [given the soup had attained a temperature of 1000C]. Substituting in (2) above, 100 = C ………………………………………………..(3)
(2) and (3) => T = kTt – 5kt + 100………………………………………………………….(4)
Now. Given the soup’s temperature is 600C in 10 minutes, substituting in (4),
60 = k{(60 x 10) – (5 x 10)} + 100 or
k = - 4/55 ANSWER ………………….……………………………………………………(5)
Part (c)
Given that the fridge can accommodate only 200C, we have from (4) and (5)
20 = (- 4/55)(20 x t) – 5(- 4/55)t + 100 or
- 80 = (- 4/55)t(20 - 5)} or
t = (80 x 55)/(4 x 15) or
t = 220/3 = 73 minutes, 20 seconds ANSWER