Temperature

Applications of Differential Equations

Temperature

An object which has a temperature of 100C is placed into air at 20C. Its temperature drops to 50C in 10 minutes. Express T(t).

Summarize the known information.

      t = 0    T = 100C

        t = 10     T = 50C

        Tm= 20C

         dT/dt = k (T – Tm)

    What do you do next?

    Plug Tm= 20C into the equation and separate the variables.

    dT/dt = k (T – 20)

    dT/(T – 20) = k dt

Now what?

    Integrate and solve for constants.

    ln(T – 20) = kt + C

    T – 20 = Cekt

    Using     t = 0     T = 100C

    C = 80

    Using t = 10    T = 50C

    k = – 0.981

The equation becomes:

    T(t) = 80e-0.0981t + 20

On to next example…

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