physics

the temperature of a body falls from 30C to 20C in 5 minutes. The air temperature is 13C. Find the temperature after a further 5 minutes.

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  1. T(s) = 13C is the air temperature,
    the initial temperature T(o) = 30C.
    Newton’s law of cooling is differential equation
    dT/dt = - k(T-T(s))
    Its solution is
    T =T(s) + (T(o) - T(s)) •e^-kt.
    After the first 5 min
    20 = T(s) + (30 – T(s)) •e^-5•k. (1)
    After a further 5 min, therefore, after 10 minutes
    T = T(s) + (30 – T(s)) •e^-10•k. (2)
    From the equation (1)
    e^-5•k = (20 -T(s))/(30-T(s))
    From the equation (2)
    e^-10•k = (e^-5•k)^2 = [(20 -T(s))/(30-T(s))]^2.
    Then the equation (2) is
    T = T(s) + (30 – T(s)) •e^-10•k =
    = T(s) + (30 – T(s)) • [(20 -T(s))/(30-T(s))]^2=
    = T(s) + (20 –T(s))^2/(30-T(s) =

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    posted by Elena

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