A coffee with temperature $LaTeX: \displaystyle 175^\circ$ is left in a room with temperature $LaTeX: \displaystyle 71^\circ$ . After 4 minutes the temperature of the coffee is $LaTeX: \displaystyle 164^\circ$ , what is the temperature of the coffee after 9 minutes?

Newton's law of Cooling states that the change in temperature is directly proportional to the difference between the object's temperature and its surroundings. $LaTeX: \frac{dT}{dt} = k(T(t)-T_{\text{room}})$ Using the substitution $LaTeX: \displaystyle y(t)=T(t)-71$ and calculating the derivative gives $LaTeX: \displaystyle \frac{dy}{dt}=\frac{dT}{dt}$ . Calculating the new initial condition using the point $LaTeX: \displaystyle (4, 164)$ and the substition gives $LaTeX: \displaystyle y(0) = T(0)-71 = 104$ . The point $LaTeX: \displaystyle (4, 164)$ must also be transformed to get $LaTeX: \displaystyle y(4) = T(4)-71 = 164 - 71 = 93$ . Substituting both of these into the equation gives the new equaiton $LaTeX: \displaystyle \frac{dy}{dt}=ky$ which has the solution $LaTeX: \displaystyle y(t) = y(0)e^{kt}=104e^{kt}$ . Evaluating the function at the point gives $LaTeX: \displaystyle 93=104e^{4k}$ and isolating the exponential gives $LaTeX: \displaystyle \frac{93}{104}=e^{4k}$ . Solving for $LaTeX: \displaystyle k$ gives $LaTeX: \displaystyle k=\frac{\ln{\left(\frac{93}{104} \right)}}{4}$ . Substuting $LaTeX: \displaystyle k$ back into the equation gives $LaTeX: \displaystyle y(t) = 104e^{\frac{\ln{\left(\frac{93}{104} \right)}}{4}t}$ and simplifying gives $LaTeX: \displaystyle y(t) = 104 \left(\frac{93}{104}\right)^{\frac{t}{4}}$ . Substituting out $LaTeX: \displaystyle y(t)$ gives $LaTeX: T(t)-71 = 104 \left(\frac{93}{104}\right)^{\frac{t}{4}} \implies\, T(t)= 104 \left(\frac{93}{104}\right)^{\frac{t}{4}} + 71$ Using $LaTeX: \displaystyle t = 9$ gives $LaTeX: \displaystyle T =104 \left(\frac{93}{104}\right)^{\frac{9}{4}} + 71\approx 152.00^\circ$