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Use differential equations to find the temperature of a cooling body

A body cools from 200 degrees to 120 degrees in 30 minutes in a room at a temperature of 60 degrees.

  1. Prove that the temperature of the body at time t minutes is

        \[ 60 + 140 e^{-kt} \]

    where

        \[ k = \frac{\log 7 - \log 3}{30}. \]

  2. Prove that the formula for the time t minutes it takes the body to cool to a temperature T degrees is given by

        \[ t = \frac{ \log 140 - \log(T-60)}{k}, \]

    where 60 < T \leq 200.

  3. Find the time at which the temperature of the cooling body is 90 degrees.
  4. If the room temperature is falling at a constant rate of 1 degree every 10 minutes find a formula for the temperature of the cooling body at time t. Assume the initial conditions are the same, the room is at 60 degrees when the body is at 200 degrees.

  1. Proof. From Newton’s law of cooling (example 3 on page 315 of Apostol) we know

        \[ f(t) = be^{-kt} + e^{-kt} \int_a^t k M(v) e^{ku} \, du \]

    is the unique solution to the differential equation telling us the temperature of a cooling body at time t under the conditions that M is the temperature of the room, and f(a) = b is the initial condition. In this problem we are given M(t) = 60, f(0) = 200. Therefore,

        \begin{align*}  f(t) &= 200 e^{-kt} + e^{-kt} \int_0^t 60 ke^{ku} \, du \\[9pt]  &= 200 e^{-kt} + 60 (1 -e^{-kt}) \\[9pt]  &= 60 + 140 e^{-kt}. \end{align*}

    Since we are given the additional information that f(30) = 120, we can compute k,

        \begin{align*}  f(30) = 120 = 60 + 140 e^{-kt} && \implies && \frac{3}{7} &= e^{-30k} \\[9pt]  && \implies && k &= \frac{\log 7 - \log 3}{30}. \qquad \blacksquare \end{align*}

  2. Proof. From part (a) we know

        \[ f(t) = 60 + 140 e^{-kt}. \]

    So, if f(t) = T then

        \begin{align*}  T = 60 + 140e^{-kt} && \implies && -kt &= \log \frac{T-60}{140} \\  && \implies && t &= \frac{\log 140 - \log (T-60)}{k}. \qquad\blacksquare \end{align*}

  3. We compute,

        \begin{align*}  f(t) = 90 && \implies && t &= \frac{\log 140 - \log(90-60)}{\frac{\log 7 - \log 3}{30}} \\[9pt]  && \implies && t &= 54.5 \text{ minutes}. \end{align*}

  4. If the temperature of the room falls at a rate of 1 degree every 10 minutes then instead of M(t) = 60 in the equation in part(a) we have M(t) = 60 - \frac{t}{10}. Computing as we did in part (a) with this new value for M(t) we have

        \begin{align*}  f(t) &= be^{-kt} + e^{-kt} \int_a^t k M(u) e^{ku} \, du \\[9pt]  &= 200e^{-kt} + e^{-kt} \int_0^t \left( 60 ke^{ku} - \frac{kue^{ku}}{10} \right) \, du \\[9pt]  &= 200e^{-kt} + e^{-kt} \left( 60 (e^{kt} -1) - \left( \frac{k}{10} \right) \left( \frac{te^{kt}}{k} - \frac{e^{kt}}{k^2} + \frac{1}{k^2} \right) \right) \\[9pt]  &= 140 e^{-kt} + 60 - \frac{t}{10} + \frac{1}{10k} - \frac{e^{-kt}}{10k} \\[9pt]  &= \frac{1}{10k} \left( 1 + k(600-t) + (1400k - 1) e^{-kt} \right). \end{align*}

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