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28 January, 20:25

Estabon poured himself a hot beverage that had a temperature of 198 then set it on the kitchen table to cool. The temperature of the kitchen was a constant 68. Of the drink cooled to 182 in five minutes how long will it take for the drink to cool to 90?

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  1. 28 January, 23:52
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    Simple sol'n:

    Rate of decrease in temperature: - 16/5 °C/min

    182 + x (-3.2) = 90

    -3.2x = - 92

    x = 28.75 min's

    Total time to cool to 90 °C = 28.75 + 5 = 33.75 min's

    By Differential Equations:

    dT/dt = - kT, t = time, T = temperature of beverage

    If you don't understand what I've written above, it is simply:

    'The change in T with respect to t is proportional to T'

    or 'The rate of cooling is proportional to T'

    Continuing:

    Separating the variables:

    1/T dT = - k dt

    Integrate both sides:

    ∫1/T dT = ∫-k dt

    ln (T) = - kt + c

    From what we're given: when t = 0, T = 198

    ln (198) = - k (0) + c

    c = ln (198)

    so ...

    ln (T) = - kt + ln (198)

    Finding k:

    ln (182) = - k (5) + ln (198)

    -5k = ln (182/198) [ln (a) - ln (b) = ln (a/b) ]

    k = - (ln (182/198)) / 5 = 0.01685 ...

    Finding desired t value:

    ln (T) = - kt + ln (198)

    ln (90) = - (0.01685 ...) t + ln (198)

    -0.01685 ... t = ln (90/198)

    t = ln (5/11) / - 0.01685 ...

    t = 46.79 min's

    Total time = 46.79 + 5 = 51.79 min's
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