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 degrees to 130 degrees Fahrenheit and keep the liquid between 110 degrees and 130 degrees as long as possible. The restaurant has three containers to select from. 1. The Centikeeper Company has a container that reduces the temperature of liquid from 200 degrees to 100 degrees in 30 minutes by maintaining a constant temperature of 70 degrees. 2. The TempControl Company has a container that reduces the temperature of a liquid from 200 degrees to 110 degrees in 25 minutes by maintaining a constant temperature of 60 degrees. 3. The Hot'N'Cold Company has a container that reduces the temperature of a liquid from 200 degrees to 120 degrees in 20 minutes by maintaining a constant temperature of 65 degrees. a) Use Newton's Law of Cooling to find a model relating the temperature of the liquid over time for watch container. b) How long does it take each container to lower the coffee temperature from 200 degrees to 130 degrees c) How long will the coffee temperature remain between 110 degrees and 130 degrees? This temperature is considered the optimal drinking temperature. e) Which company would you recommend to the restaurant? Why?

Explanation / Answer

Use the equation U(t) = T+(U1-T)(Kt)

1.Initial heat, U1 = 200 degrees and T = 70
At time t = 30 minutes, the heat U = 100

U(t) = T+(U1-T)(Kt)

100= 700+(200-700)(30K)

13030K = 30

K=ln(30/130)/30 = -0.04887

2.In second case, U1 = 200 and T = 60degrees
At the time t = 25, the heat U(t) = 110 degrees

      U(t) = T+(U1-T)(Kt)

     110= 600+(200-600)(25K)

     14025K   = 50

     K=ln(50/140)/25 = -0.04118

3.The initial heat U1 = 200 and T = 65
At the time, t = 20, the heat U = 120

     U(t) = T+(U1-T)(Kt)

     120= 650+(200-650)(20K)

     13520K   = 55

     K= ln(55/135)/20 = -0.04489

(a)

Substitute values of K in equation U(t) = T+(U1-T)(Kt)at particular temperatures.

U(t) = 700+130(-0.04887t)

U(t) = 600+140(-0.04118t)

U(t) = 650+135(-0.04489t)

(b)

Find out the time to reach the temperature 130 degrees.

T1 = ln(60/130)/(-0.04887) = 15.82

T2 = ln(70/140)/(-0.04118) = 16.83

T3 = ln(65/135)/(-0.04489) = 16.28

(c)

Find the time to reach 110 degrees.

T1 = ln(40/130)/(-0.04887) = 24.11

T1 = ln(50/140)/(-0.04118) = 25.00

T1 = ln(45/135)/(-0.04489) = 24.47

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