We take a milk carton from the refrigerator and put it on the table at room temperature. We assume that the temperature of the carton, T, as

Question

We take a milk carton from the refrigerator and put it on the table at room temperature. We assume that the temperature of the carton, T, as a function of time follows Newton’s cooling law:
dT / dt = k (T – a),
where a is the temperature of the surroundings and k is a constant. We assume the temperature
the surroundings are a = 20 (given in ◦C), and that T (0) = 4 (given in ◦C).
a) Show that T (t) = 20 – 16e ^ kt (given in ◦C).
b) After 5 minutes the temperature of the carton is 8◦C. What is the temperature after 15 minutes?

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Answers ( No )

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    2021-11-23T22:36:48+00:00

    We can solve this problem using separation of variables.

    Then apply the initial conditions

    EXPLANATION

    We were given the first order differential equation

    \frac{dT}{dt}=k(T-a)

    We now separate the time and the temperature variables as follows,

    \frac{dT}{T-a}=kdt

    Integrating both sides of the differential equation, we obtain;

    ln(T-a)=kt +c

    This natural logarithmic equation can be rewritten as;

    T-a=e^{kt +c}

    Applying the laws of exponents, we obtain,

    T-a=e^{kt}\times e^{c}

    T-a=e^{c}e^{kt}

    We were given the initial conditions,

    T(0)=4

    Let us apply this condition to obtain;

    4-20=e^{c}e^{k(0)}

    -16=e^{c}

    Now our equation, becomes

    T-a=-16e^{kt}

    or

    T=a-16e^{kt}

    When we substitute a=20,
    we obtain,

    T=20-16e^{kt}

    b) We were also given that,

    T(5)=8

    Let us apply this condition again to find k.

    8=20-16e^{5k}

    This implied

    -12=-16e^{5k}

    \frac{-12}{-16}=e^{5k}

    \frac{3}{4}=e^{5k}

    We take logarithm to base e of both sides,

    ln(\frac{3}{4})=5k

    This implies that,

    \frac{ln(\frac{3}{4})}{5}=k

    k=-0.2877

    After 15 minutes, the temperature will be,

    T=20-16e^{-0.2877\times 15}

    T=20-0.21376

    T=19.786

    After 15 minutes, the temperature is approximately 20°C

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