A fast-food restaurant wants a special container to hold coffee. The restaurant

Question

A fast-food restaurant wants a special container to hold coffee. The restaurant wishes the container to quickly cool the coffee from 200 degree to 130 degree F and keep the liquid between 110 degree and 130 degree F as long as possible. The restaurant has three containers to select from. 1. The CentiKeeper Company has a container that reduces the temperature of a liquid from 200 degree to 100 degree F in 30 minutes by maintaining a constant temperature of 70 degree F. 2. The TempControl Company has a container that reduces the temperature of a liquid from 200 degree to 110 degree F in 25 minutes by maintaining a constant temperature of 60 degree F. 3. The Hot?n?Cold Company has a container that reduces the temperature of a liquid from 200 degree to 120 degree F in 20 minutes by maintaining a constant temperature of 65 degree F. You need to recommend which container the restaurant should purchase. (a) Use Newton's Law of Cooling to find a model relating the temperature o

Explanation / Answer

Newton’s Law of Cooling:

U (t) = T + (U0 – T) e^kt

T = Ambient temperature (constant temperature surrounding object)
U0 = Initial temperature of cooling item
K= Negative constant
t= Time

Container 1: The CentiKeeper company

Container 2: The Tempcontrol company

Container 3: The Hot'n'Cold company

Let us do each container in turn using U (t) = T + (U0 – T) e^kt and the given data to find K.

1.
U0 = 200 and T = 70
At t = 30 we have U = 100

100 = 70 + (200 - 70)*e^(K*30)
30 = 130*e^(K*30)
K*30 = ln(30/130) = ln(3/13) and K = ln(3/13)/30 = -0.048877902

2.
U0 = 200 and T = 60
At t = 25 we have U = 110

As above gives: 110 = 60 + (200 - 60)*e^(K*25)
K = ln(50/140)/25 = ln(5/14)/20 = -0.041184777

3.
U0 = 200 and T = 65
At t = 20 we have U = 120

120 = 65 + (200 - 65)*e^(K*20)
K = ln(55/135)/20 = -0.04489708

a)

Using Newton’s Law of Cooling:

Container 1: U(t) = 70 + 130*e^(-0.048877902*t)
Container 2: U(t) = 60 + 140*e^(-0.041184777*t)
Container 3: U(t) = 65 + 135*e^( -0.04489708*t)

b)

Time to reach 130. Solve each equation for t for a temperature of 130.
Container 1: t = ln(60/130)/(-0.048877902) = 15.81880
Container 2: t = ln(70/140)/(-0.041184777) = 16.83018
Container 3: t= ln(65/135)/(-0.041184777) = 16.27918

c)

First find the time to reach 110 and then take the difference with part B results.
Container 1: t = ln(40/130)/(-0.048877902) = 24.11427
Container 2: t = ln(50/140)/(-0.041184777) = 25
Container 3: t= ln(45/135)/(-0.041184777) = 24.46957

Time the temperature will be between 110 and 130.
Container 1: t = 24.11427 - 15.81880 = 8.29547
Container 2: t = 25 - 16.83018 = 8.16982
Container 3: t = 24.46957 - 16.27918 = 8.19039

e)

It is recommended to use the CentiKeeper Company because compared to the other companies it cools down the faster while maintaining its temperature the longest.

f)

The cost of the container greatly affects the decision to go with one company over the other because there are only slight differences between the cups. All three companies cool down the contents of the cup in less than 17 minutes and all three of them hold the ideal temperature for more than 8 minutes. When it comes to recommending a company over another if I had the cost of each cup I would have recommended the cheapest one.

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