A coffee with temperature $LaTeX: \displaystyle 171^\circ$ is left in a room with temperature $LaTeX: \displaystyle 55^\circ$ . After 5 minutes the temperature of the coffee is $LaTeX: \displaystyle 165^\circ$ , what is the temperature of the coffee after 14 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 $LaTeX: \displaystyle y(t)=T(t)-55$ and calculating the derivative gives $LaTeX: \displaystyle \frac{dy}{dt}=\frac{dT}{dt}$ . Calculating the new initial condition using the point $LaTeX: \displaystyle (5, 165)$ and the substition gives $LaTeX: \displaystyle y(0) = T(0)-55 = 116$ . The point $LaTeX: \displaystyle (5, 165)$ must also be transformed to get $LaTeX: \displaystyle y(5) = T(5)-55 = 165 - 55 = 110$ . Substituting both of these into the equation gives the new equaiton $LaTeX: \displaystyle \frac{dy}{dt}=ky$ which has the solution $LaTeX: \displaystyle y(t) = y(0)e^{kt}=116e^{kt}$ . Evaluating the function at the point gives $LaTeX: \displaystyle 110=116e^{5k}$ and isolating the exponential gives $LaTeX: \displaystyle \frac{55}{58}=e^{5k}$ . Solving for $LaTeX: \displaystyle k$ gives $LaTeX: \displaystyle k=\frac{\ln{\left(\frac{55}{58} \right)}}{5}$ . Substuting $LaTeX: \displaystyle k$ back into the equation gives $LaTeX: \displaystyle y(t) = 116e^{\frac{\ln{\left(\frac{55}{58} \right)}}{5}t}$ and simplifying gives $LaTeX: \displaystyle y(t) = 116 \left(\frac{55}{58}\right)^{\frac{t}{5}}$ . Substituting out $LaTeX: \displaystyle y(t)$ gives $LaTeX: T(t)-55 = 116 \left(\frac{55}{58}\right)^{\frac{t}{5}} \implies\, T(t)= 116 \left(\frac{55}{58}\right)^{\frac{t}{5}} + 55$ Using $LaTeX: \displaystyle t = 14$ gives $LaTeX: \displaystyle T =116 \left(\frac{55}{58}\right)^{\frac{14}{5}} + 55\approx 155.00^\circ$