The formula below, Newton's law of cooling, allows us to calculate the temperature of an object as it cools. T = R + (T0 - R) e-kt T = temperature of the object after t minutes T0 = initial temperature R = temperature of the surroundings k = cooling rate (depends on the object and conditions) Suppose the temperature in Denver, Colorado was 55°F when a 5°F arctic cold front moved over the state. Using k = 0.012, how long would it take a puddle of water to freeze? (Recall that water freezes at 32°F.)

Question

Makena Holloway

Mathematics

2 October, 15:18

The formula below, Newton's law of cooling, allows us to calculate the temperature of an object as it cools. T = R + (T0 - R) e-kt T = temperature of the object after t minutes T0 = initial temperature R = temperature of the surroundings k = cooling rate (depends on the object and conditions) Suppose the temperature in Denver, Colorado was 55°F when a 5°F arctic cold front moved over the state. Using k = 0.012, how long would it take a puddle of water to freeze? (Recall that water freezes at 32°F.)

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Beau Farrell · Answer 1

The Newton's law of cooling:

T = R + (To - R) * e ^ (-kt)

We know that: T = 32° F (the water freezes at 32° F), To = 55° F, R = 5° F

and the cooling rate is: k = 0.012.

33° F = 5° F + (55° F - 5° F) · 2.72^ (-0.012 t)

33 - 5 = 50 · 2.72^ (-0.012 t)

27 = 50 · 2.72^ (-0.012 t)

2.72^ (-0.012 t) = 27/50

1 / 2.72^ (0.012 t) = 0.54

2.72^ (0.012 t) = 1 : 0.54 = 1.85

0.012 t = log (base e) 1.85 = ln 1.85 = 0.6

t = 0.6 : 0.012 = 50

Answer: It would take 50 minutes.