A coffee with temperature $LaTeX: \displaystyle 169^\circ$ is left in a room with temperature $LaTeX: \displaystyle 73^\circ$ . After 10 minutes the temperature of the coffee is $LaTeX: \displaystyle 134^\circ$ , how long until the coffee is $LaTeX: \displaystyle 112^\circ$ ?

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 $LaTeX: \displaystyle y(t)=T(t)-73$ and calculating the derivative gives $LaTeX: \displaystyle \frac{dy}{dt}=\frac{dT}{dt}$ . Calculating the new initial condition using the point $LaTeX: \displaystyle (10, 134)$ and the substition gives $LaTeX: \displaystyle y(0) = T(0)-73 = 96$ . The point $LaTeX: \displaystyle (10, 134)$ must also be transformed to get $LaTeX: \displaystyle y(10) = T(10)-73 = 134 - 73 = 61$ . Substituting both of these into the equation gives the new equaiton $LaTeX: \displaystyle \frac{dy}{dt}=ky$ which has the solution $LaTeX: \displaystyle y(t) = y(0)e^{kt}=96e^{kt}$ . Evaluating the function at the point gives $LaTeX: \displaystyle 61=96e^{10k}$ and isolating the exponential gives $LaTeX: \displaystyle \frac{61}{96}=e^{10k}$ . Solving for $LaTeX: \displaystyle k$ gives $LaTeX: \displaystyle k=\frac{\ln{\left(\frac{61}{96} \right)}}{10}$ . Substuting $LaTeX: \displaystyle k$ back into the equation gives $LaTeX: \displaystyle y(t) = 96e^{\frac{\ln{\left(\frac{61}{96} \right)}}{10}t}$ and simplifying gives $LaTeX: \displaystyle y(t) = 96 \left(\frac{61}{96}\right)^{\frac{t}{10}}$ . Substituting out $LaTeX: \displaystyle y(t)$ gives $LaTeX: T(t)-73 = 96 \left(\frac{61}{96}\right)^{\frac{t}{10}} \implies\, T(t)= 96 \left(\frac{61}{96}\right)^{\frac{t}{10}} + 73$ Using $LaTeX: \displaystyle T$ gives the equation $LaTeX: \displaystyle 112=96 \left(\frac{61}{96}\right)^{\frac{t}{10}} + 73$ . Isolating the exponential gives $LaTeX: \displaystyle \frac{13}{32}=\left(\frac{61}{96}\right)^{\frac{t}{10}}$ . Taking the natural logarithm of both sides and solving for $LaTeX: \displaystyle t$ gives $LaTeX: \displaystyle t = \frac{10 \ln{\left(\frac{13}{32} \right)}}{\ln{\left(\frac{61}{96} \right)}}\approx 19.9$ minutes.