A coffee with temperature $LaTeX: \displaystyle 162^\circ$ is left in a room with temperature $LaTeX: \displaystyle 56^\circ$ . After 15 minutes the temperature of the coffee is $LaTeX: \displaystyle 156^\circ$ , how long until the coffee is $LaTeX: \displaystyle 140^\circ$ ?

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 $LaTeX: \displaystyle y(t)=T(t)-56$ and calculating the derivative gives $LaTeX: \displaystyle \frac{dy}{dt}=\frac{dT}{dt}$ . Calculating the new initial condition using the point $LaTeX: \displaystyle (15, 156)$ and the substition gives $LaTeX: \displaystyle y(0) = T(0)-56 = 106$ . The point $LaTeX: \displaystyle (15, 156)$ must also be transformed to get $LaTeX: \displaystyle y(15) = T(15)-56 = 156 - 56 = 100$ . Substituting both of these into the equation gives the new equaiton $LaTeX: \displaystyle \frac{dy}{dt}=ky$ which has the solution $LaTeX: \displaystyle y(t) = y(0)e^{kt}=106e^{kt}$ . Evaluating the function at the point gives $LaTeX: \displaystyle 100=106e^{15k}$ and isolating the exponential gives $LaTeX: \displaystyle \frac{50}{53}=e^{15k}$ . Solving for $LaTeX: \displaystyle k$ gives $LaTeX: \displaystyle k=\frac{\ln{\left(\frac{50}{53} \right)}}{15}$ . Substuting $LaTeX: \displaystyle k$ back into the equation gives $LaTeX: \displaystyle y(t) = 106e^{\frac{\ln{\left(\frac{50}{53} \right)}}{15}t}$ and simplifying gives $LaTeX: \displaystyle y(t) = 106 \left(\frac{50}{53}\right)^{\frac{t}{15}}$ . Substituting out $LaTeX: \displaystyle y(t)$ gives $LaTeX: T(t)-56 = 106 \left(\frac{50}{53}\right)^{\frac{t}{15}} \implies\, T(t)= 106 \left(\frac{50}{53}\right)^{\frac{t}{15}} + 56$ Using $LaTeX: \displaystyle T$ gives the equation $LaTeX: \displaystyle 140=106 \left(\frac{50}{53}\right)^{\frac{t}{15}} + 56$ . Isolating the exponential gives $LaTeX: \displaystyle \frac{42}{53}=\left(\frac{50}{53}\right)^{\frac{t}{15}}$ . Taking the natural logarithm of both sides and solving for $LaTeX: \displaystyle t$ gives $LaTeX: \displaystyle t = \frac{15 \ln{\left(\frac{42}{53} \right)}}{\ln{\left(\frac{50}{53} \right)}}\approx 59.9$ minutes.