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