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    Topics || Problems

    A thermometer reading is \(18^oF\) is brought into a room where the temperature is \(70^oF\); 1 min later the thermometer reading is \(31^oF\). Determine the temperature reading as a function of time and, in particular, find the temperature reading 5 min after the thermometer is first brought into the room.
    Room Temperature \(70^oF\)
    Time Thermometer Reading
    \(0 ~min\) \(18^oF\)
    \(1~min\) \(31^oF\)

    Temperature reading as a function of time.

    Fom Newtons Law of Cooling \(\frac{dT_b}{dt} = k (T_b - T_s) \), where \(T_b\) is the temperature of the body, \(T_s\) is the temperature of the sorroundings and k is a constant.

    \(\frac{dT_b}{T_b-T_s} = kdt\), \(T_s\) is the room temperature and assumed to be constant.

    \(\ln (T_b - T_s) + ln C= kt \)

    At \(t=0 ~T_s = 70^oF, ~ T_b = 18^oF\)

    \(\ln (18-70) + ln C= 0\)

    \(\ln C = -ln (-52)\)

    \(\ln(T_b - T_s) - \ln(-52) = kt\)

    At \(t = 1~ T_s = 31^oF\)

    \(\ln(31-70)-\ln(-52) = k\)

    \(k = \ln \frac{-39}{-52}\)

    \(k = \ln \frac{3}{4} = -0.29\)

    \(\ln(T_b - 70) - \ln(-52) = \ln (\frac{3}{4})t\)

    \(\frac{T_b-70}{-52} = e^{-0.29t}\)

    \(T_b = 70 -52 e^{-0.29t}\) Answer

    At \(t = 5\)

    \(T_b = 70-52e^(-.29x5)\)

    \(T_b \approx 57.66^oF \approx 58^oF\) Answer