Answer to Question #316650 in Differential Equations for Tobias Felix

Question

1.At 1:00 P.M., a thermometer reading 70°F is taken outside where the air temperature is -10°F. At 1:02 P.M., the reading is 26°F. At 1:05 P.M., the thermometer is taken back indoors, where the air is at 70°F. What is the temperature reading at 1:09 P.M.?

Answer 1

$T-temperatrure\\T_{out}-outside\\\frac{dT}{dt}=k\left( T_{out}-T \right) ,T\left( t_0 \right) =T_0\Rightarrow T\left( t \right) =T_{out}+\left( T_0-T_{out} \right) e^{-k\left( t-t_0 \right)}=\left[ e^{-k}=A \right] =T_{out}+\left( T_0-T_{out} \right) A^{t-t_0}\\\\T_{out}=-10, t_0=0,T_0=70,T\left( 2 \right) =26\Rightarrow -10+\left( 70+10 \right) A^{2-0}=26\Rightarrow A=\sqrt{\frac{36}{80}}=0.67082\\At\,\,t=5: T\left( 5 \right) =-10+80\cdot A^{5-0}=-10+80\cdot 0.67082^5=0.867259\\T_{out}=70,t_0=5,T_0=0.867259\Rightarrow T\left( 9 \right) =70+\left( 0.867259-70 \right) \cdot 0.67082^{9-5}=56.0007$