A coffee with temperature $LaTeX: \displaystyle 150^\circ$ is left in a room with temperature $LaTeX: \displaystyle 60^\circ$ . After 6 minutes the temperature of the coffee is $LaTeX: \displaystyle 140^\circ$ , how long until the coffee is $LaTeX: \displaystyle 118^\circ$ ?

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 $LaTeX: \displaystyle y(t)=T(t)-60$ and calculating the derivative gives $LaTeX: \displaystyle \frac{dy}{dt}=\frac{dT}{dt}$ . Calculating the new initial condition using the point $LaTeX: \displaystyle (6, 140)$ and the substition gives $LaTeX: \displaystyle y(0) = T(0)-60 = 90$ . The point $LaTeX: \displaystyle (6, 140)$ must also be transformed to get $LaTeX: \displaystyle y(6) = T(6)-60 = 140 - 60 = 80$ . Substituting both of these into the equation gives the new equaiton $LaTeX: \displaystyle \frac{dy}{dt}=ky$ which has the solution $LaTeX: \displaystyle y(t) = y(0)e^{kt}=90e^{kt}$ . Evaluating the function at the point gives $LaTeX: \displaystyle 80=90e^{6k}$ and isolating the exponential gives $LaTeX: \displaystyle \frac{8}{9}=e^{6k}$ . Solving for $LaTeX: \displaystyle k$ gives $LaTeX: \displaystyle k=\frac{\ln{\left(\frac{8}{9} \right)}}{6}$ . Substuting $LaTeX: \displaystyle k$ back into the equation gives $LaTeX: \displaystyle y(t) = 90e^{\frac{\ln{\left(\frac{8}{9} \right)}}{6}t}$ and simplifying gives $LaTeX: \displaystyle y(t) = 90 \left(\frac{8}{9}\right)^{\frac{t}{6}}$ . Substituting out $LaTeX: \displaystyle y(t)$ gives $LaTeX: T(t)-60 = 90 \left(\frac{8}{9}\right)^{\frac{t}{6}} \implies\, T(t)= 90 \left(\frac{8}{9}\right)^{\frac{t}{6}} + 60$ Using $LaTeX: \displaystyle T$ gives the equation $LaTeX: \displaystyle 118=90 \left(\frac{8}{9}\right)^{\frac{t}{6}} + 60$ . Isolating the exponential gives $LaTeX: \displaystyle \frac{29}{45}=\left(\frac{8}{9}\right)^{\frac{t}{6}}$ . Taking the natural logarithm of both sides and solving for $LaTeX: \displaystyle t$ gives $LaTeX: \displaystyle t = \frac{6 \ln{\left(\frac{29}{45} \right)}}{\ln{\left(\frac{8}{9} \right)}}\approx 22.4$ minutes.