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27 October, 04:46

A cake is removed from a 375°F oven and placed on a cooling rack in a 63°F room. After 30 minutes the cake is 175°F. When will it be 150°F? (Round your answer to two decimal places.)

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  1. 27 October, 06:36
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    The cake will be at temperature 150°F at after 37.34 minutes

    Explanation:

    Let T be the temperature of the cake at any time

    T∞ be the temperature around the cooling rack = 63°F

    T₀ be the initial temperature of the cake = 375°F

    And m, c, h are all constants from the cooling law relation

    From Newton's law of cooling

    Rate of Heat loss by the cake = Rate of Heat gain by the environment

    - mc (d/dt) (T - T∞) = h (T - T∞)

    (d/dt) (T - T∞) = dT/dt (Because T∞ is a constant)

    dT/dt = (-h/mc) (T - T∞)

    Let (h/mc) be k

    dT / (T - T∞) = - kdt

    Integrating the left hand side from T₀ to T and the right hand side from 0 to t

    In [ (T - T∞) / (T₀ - T∞) ] = - kt

    (T - T∞) / (T₀ - T∞) = e⁻ᵏᵗ

    (T - T∞) = (T₀ - T∞) e⁻ᵏᵗ

    Inserting the known variables

    (T - 63) = (375 - 63) e⁻ᵏᵗ

    (T - 63) = 312 e⁻ᵏᵗ

    At 30 minutes, T = 175°F

    175 - 63 = 312 e⁻ᵏᵗ

    112/312 = e⁻ᵏᵗ

    - kt = In (112/312) = In (0.3590)

    - 30k = - 1.025

    k = 1.025/30 = 0.0342 / min

    When the temp is 150°F,

    (T - T∞) = (T₀ - T∞) e⁻ᵏᵗ

    (150 - 63) = 312 e⁻ᵏᵗ

    e⁻ᵏᵗ = (87/312) = 0.2788

    - kt = In 0.2788 = - 1.277

    t = 1.277/k = 1.277/0.0342 = 37.34 min.
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