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24 February, 14:30

5. Suppose a cup of cocoa cooled from 90° C to 60° C after 10 minutes in a room whose temperature was 20° C. Use Newton's law of cooling [T - Ts = (T0 - Ts) ⋅ e-k⋅t ] to answer the following questions: a) How much longer would it take the cocoa to cool to 35° C? b) Instead of being left to stand in a room, the cup with initial temperature 90° C is placed in a freezer whose temperature is - 15° C. How long will it take the cocoa to cool from 90° C to 35° C?

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  1. 24 February, 17:49
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    a) t = 1051.6 sec = 17.5 min

    b) t = 795.5 sec = 13.25 min

    Explanation:

    First of all we use the initial data to find out constant 'K'.

    T - Ts = (T₀ - Ts) e^ (-kt)

    Here, we have:

    T = Final Temperature = 60° C

    Ts = Surrounding Temperature = 20° C

    T₀ = Initial Temperature = 90° C

    t = time = 10 min = 600 sec

    k = constant = ?

    Therefore,

    60° C - 20° C = (90° C - 20° C). e^ (-k600)

    40° C/70° C = e^ (-k600)

    ln (0.57142) = - 600k

    k = 9.327 x 10⁻⁴ sec⁻¹

    a)

    Now, for this case we have:

    T = Final Temperature = 35° C

    Ts = Surrounding Temperature = 20° C

    T₀ = Initial Temperature = 60° C

    t = time = ?

    k = constant = 9.327 x 10⁻⁴ sec⁻¹

    Therefore,

    35° C - 20° C = (60° C - 20° C). e^ (-9.327 x 10⁻⁴ sec⁻¹ x t)

    15° C/40° C = e^ (-9.327 x 10⁻⁴ sec⁻¹ x t)

    ln (15/40) = - 9.327 x 10⁻⁴ sec⁻¹ x t

    t = 1051.6 sec = 17.5 min

    b)

    Now, for this case we have:

    T = Final Temperature = 35° C

    Ts = Surrounding Temperature = - 15° C

    T₀ = Initial Temperature = 90° C

    t = time = ?

    k = constant = 9.327 x 10⁻⁴ sec⁻¹

    Therefore,

    35° C + 15° C = (90° C + 15° C). e^ (-9.327 x 10⁻⁴ sec⁻¹ x t)

    50° C/105° C = e^ (-9.327 x 10⁻⁴ sec⁻¹ x t)

    ln (50/105) = - 9.327 x 10⁻⁴ sec⁻¹ x t

    t = 795.5 sec = 13.25 min
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