A coffee with temperature $LaTeX: \displaystyle 175^\circ$ is left in a room with temperature $LaTeX: \displaystyle 77^\circ$ . After 12 minutes the temperature of the coffee is $LaTeX: \displaystyle 157^\circ$ , how long until the coffee is $LaTeX: \displaystyle 149^\circ$ ?

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 $LaTeX: \displaystyle y(t)=T(t)-77$ and calculating the derivative gives $LaTeX: \displaystyle \frac{dy}{dt}=\frac{dT}{dt}$ . Calculating the new initial condition using the point $LaTeX: \displaystyle (12, 157)$ and the substition gives $LaTeX: \displaystyle y(0) = T(0)-77 = 98$ . The point $LaTeX: \displaystyle (12, 157)$ must also be transformed to get $LaTeX: \displaystyle y(12) = T(12)-77 = 157 - 77 = 80$ . Substituting both of these into the equation gives the new equaiton $LaTeX: \displaystyle \frac{dy}{dt}=ky$ which has the solution $LaTeX: \displaystyle y(t) = y(0)e^{kt}=98e^{kt}$ . Evaluating the function at the point gives $LaTeX: \displaystyle 80=98e^{12k}$ and isolating the exponential gives $LaTeX: \displaystyle \frac{40}{49}=e^{12k}$ . Solving for $LaTeX: \displaystyle k$ gives $LaTeX: \displaystyle k=\frac{\ln{\left(\frac{40}{49} \right)}}{12}$ . Substuting $LaTeX: \displaystyle k$ back into the equation gives $LaTeX: \displaystyle y(t) = 98e^{\frac{\ln{\left(\frac{40}{49} \right)}}{12}t}$ and simplifying gives $LaTeX: \displaystyle y(t) = 98 \left(\frac{40}{49}\right)^{\frac{t}{12}}$ . Substituting out $LaTeX: \displaystyle y(t)$ gives $LaTeX: T(t)-77 = 98 \left(\frac{40}{49}\right)^{\frac{t}{12}} \implies\, T(t)= 98 \left(\frac{40}{49}\right)^{\frac{t}{12}} + 77$ Using $LaTeX: \displaystyle T$ gives the equation $LaTeX: \displaystyle 149=98 \left(\frac{40}{49}\right)^{\frac{t}{12}} + 77$ . Isolating the exponential gives $LaTeX: \displaystyle \frac{36}{49}=\left(\frac{40}{49}\right)^{\frac{t}{12}}$ . Taking the natural logarithm of both sides and solving for $LaTeX: \displaystyle t$ gives $LaTeX: \displaystyle t = \frac{12 \ln{\left(\frac{36}{49} \right)}}{\ln{\left(\frac{40}{49} \right)}}\approx 18.2$ minutes.