A pizza pan is removed at 9:00 PM from an oven whose temperature is fixed at 450°F into a room that is a constant 70°F. After 5​ minutes, th

Question

A pizza pan is removed at 9:00 PM from an oven whose temperature is fixed at 450°F into a room that is a constant 70°F. After 5​ minutes, the pizza pan is at 300°F.
​(a) At what time is the temperature of the pan 125°F? ​
(b) Determine the time that needs to elapse before the pan is 150°. ​
(c) What do you notice about the temperature as time​ passes?

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Latifah 1 week 2021-07-22T08:26:45+00:00 1 Answers 2 views 0

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    2021-07-22T08:28:28+00:00

    Answer:

    A) It will get to a temperature of 125°F at 9:19 PM

    B) It will get to a temperature of 150°F at 9:16 PM

    C) as time passes temperature approaches the initial temperature of 450°F

    Step-by-step explanation:

    We are given;

    Initial temperature; T_i = 450°F

    Room temperature; T_r = 70°F

    From Newton’s law of cooling, temperature after time (t) is given as;

    T(t) = T_r + (T_i – T_r)e^(-kt)

    Where k is cooling rate and t is time after the initial temperature.

    Now, we are told that After 5​ minutes, the temperature is 300°F.

    Thus;

    300 = 70 + (450 – 70)e^(-5k)

    300 – 70 = 380e^(-5k)

    230/380 = e^(-5k)

    e^(-5k) = 0.6053

    -5k = In 0.6053

    -5k = -0.502

    k = 0.502/5

    k = 0.1004 /min

    A) Thus, at temperature of 125°F, we can find the time from;

    125 = 70 + (450 – 70)e^(-0.1004t)

    125 – 70 = 380e^(-0.1004t)

    55/380 = e^(-0.1004t)

    In (55/380) = -0.1004t

    -0.1004t = -1.9328

    t = 1.9328/0.1018

    t ≈ 19 minutes

    Thus, it will get to a temperature of 125°F at 9:19 PM

    B) Thus, at temperature of 150°F, we can find the time from;

    150 = 70 + (450 – 70)e^(-0.1004t)

    150 – 70 = 380e^(-0.1004t)

    80/380 = e^(-0.1004t)

    In (80/380) = -0.1004t

    -0.1004t = -1.5581

    t = 1.5581/0.1004

    t ≈ 16 minutes.

    Thus, it will get to a temperature of 150°F at 9:16 PM

    C) As time passes which means as it approaches to infinity, it means that e^(-kt) gets to 1.

    Thus,we have;

    T(t) = T_r + (T_i – T_r)

    T_r will cancel out to give;

    T(t) = T_i

    Thus, as time passes temperature approaches the initial temperature of 450°F

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