MATH SOLVE

4 months ago

Q:
# A metal beam was brought from the outside cold into a machine shop where the temperature was held at 70°F After 10 min, the beam warmed to 40°F and after another 10 min it was 55°F. Use Newton's Law of Cooling to estimate the beam's initial temperature The beam's initial temperature was "F

Accepted Solution

A:

Answer:The beam's initial temperature was 10°FStep-by-step explanation:Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature. This means that:[tex]\frac{dT}{dt} =-k (T-T_{a})[/tex] where k is a positive constant and [tex]T_{a}[/tex] is the ambient temperature.This is the solution of the differential equation[tex]T(t)=T_{a}+T_{0}\cdot e^{(kt)}[/tex] where [tex]T(t)[/tex] is the temperature after t minutes and [tex]T_{0}[/tex] and k are constants yet to be determined.We know from the information given that the ambient temperature is 70°F, so[tex]T(t)=70+T_{0}\cdot e^{(kt)}[/tex]We also know that [tex]T(10) = 40 \:F[/tex] and [tex]T(20) = 55 \:F[/tex], we can use these to determine the constants [tex]T_{0}[/tex] and k.If we use the first condition [tex]T(10) = 40 \:F[/tex] we have[tex]40=70+T_{0}\cdot e^{(k\cdot 10)}[/tex]We can solve for k in terms of [tex]T_{0}[/tex] as follows[tex]40=70+T_{0}\cdot e^{(k\cdot 10)}\\70+T_0e^{k\cdot 10}=40\\T_0e^{k \cdot 10}=-30\\e^{k \cdot 10}=-\frac{30}{T_0}\\\ln \left(e^{k\cdot \:10}\right)=\ln \left(-\frac{30}{T_0}\right)\\k\cdot \:10\ln \left(e\right)=\ln \left(-\frac{30}{T_0}\right)\\k=\frac{\ln \left(-\frac{30}{T_0}\right)}{10}[/tex]We can rewrite [tex]T(t)[/tex] as[tex]T(t)=70+T_{0}\cdot e^{(\frac{\ln \left(-\frac{30}{T_0}\right)}{10}\cdot t)}[/tex]Next we use the second condition [tex]T(20) = 55 \:F[/tex] to get[tex]55=70+T_{0}\cdot e^{(\frac{\ln \left(-\frac{30}{T_0}\right)}{10}\cdot 20)}[/tex]and we solve for [tex]T_{0}[/tex] [tex]55=70+T_{0}\cdot e^{(\frac{\ln \left(-\frac{30}{T_0}\right)}{10}\cdot 20)}\\-15=T_{0}\cdot e^{2\ln \left(-\frac{30}{T_0}\right)}\\-15=T_{0}\cdot e^{\ln \left(-\frac{30}{T_0}\right)^{2}}\\-15=T_{0}\cdot \left(-\frac{30}{T_0}\right)^{2}\\-15=T_{0} \cdot \left(\frac{900}{T_0^2}\right)\\-15=\frac{900}{T_{0}} \\T_{0} = -60[/tex]The value of k is [tex]k=\frac{\ln \left(-\frac{30}{T_0}\right)}{10}\\k=\frac{\ln \left(\frac{-30}{-60}\right)}{10}\\k=-\frac{ln(2)}{10}[/tex]So the general solution of the equation is[tex]T(t)=70-60\cdot e^{(-\frac{ln(2)}{10}\cdot t)}[/tex]In particular, since we want to know T(0), we can now just evaluate:[tex]T(0)=70-60\cdot e^{(-\frac{ln(2)}{10}\cdot 0)}\\T(0)=10[/tex]