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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. \begin{equation*} \frac{dT}{dt} = k(T(t)-T_{\text{room}})\end{equation*}Using the substitution \(\displaystyle y(t)=T(t)-55\) and calculating the derivative gives \(\displaystyle \frac{dy}{dt}=\frac{dT}{dt}\). Calculating the new initial condition using the point \(\displaystyle (5, 165)\) and the substition gives \(\displaystyle y(0) = T(0)-55 = 116\). The point \(\displaystyle (5, 165)\) must also be transformed to get \(\displaystyle y(5) = T(5)-55 = 165 - 55 = 110\). Substituting both of these into the equation gives the new equaiton \(\displaystyle \frac{dy}{dt}=ky\) which has the solution \(\displaystyle y(t) = y(0)e^{kt}=116e^{kt}\). Evaluating the function at the point gives \(\displaystyle 110=116e^{5k}\) and isolating the exponential gives \(\displaystyle \frac{55}{58}=e^{5k}\). Solving for \(\displaystyle k\) gives \(\displaystyle k=\frac{\ln{\left(\frac{55}{58} \right)}}{5}\). Substuting \(\displaystyle k\) back into the equation gives \(\displaystyle y(t) = 116e^{\frac{\ln{\left(\frac{55}{58} \right)}}{5}t}\) and simplifying gives \(\displaystyle y(t) = 116 \left(\frac{55}{58}\right)^{\frac{t}{5}}\). Substituting out \(\displaystyle y(t)\) gives \begin{equation*}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 \end{equation*}Using \(\displaystyle t = 14\) gives \(\displaystyle T =116 \left(\frac{55}{58}\right)^{\frac{14}{5}} + 55\approx 155.00^\circ\)

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\begin{question}A coffee with temperature $171^\circ$ is left in a room with temperature $55^\circ$. After 5 minutes the temperature of the coffee is $165^\circ$, what is the temperature of the coffee after 14 minutes?
    \soln{9cm}{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. \begin{equation*} \frac{dT}{dt} = k(T(t)-T_{\text{room}})\end{equation*}Using the substitution $y(t)=T(t)-55$ and calculating the derivative gives $\frac{dy}{dt}=\frac{dT}{dt}$. Calculating the new initial condition using the point $(5, 165)$ and the substition gives $y(0) = T(0)-55 = 116$. The point $(5, 165)$ must also be transformed to get $y(5) = T(5)-55 = 165 - 55 = 110$. Substituting both of these into the equation gives the new equaiton $\frac{dy}{dt}=ky$ which has the solution $y(t) = y(0)e^{kt}=116e^{kt}$. Evaluating the function at the point gives $110=116e^{5k}$ and isolating the exponential gives $\frac{55}{58}=e^{5k}$. Solving for $k$ gives $k=\frac{\ln{\left(\frac{55}{58} \right)}}{5}$.  Substuting $k$ back into the equation gives $y(t) = 116e^{\frac{\ln{\left(\frac{55}{58} \right)}}{5}t}$ and simplifying gives $y(t) = 116 \left(\frac{55}{58}\right)^{\frac{t}{5}}$. Substituting out $y(t)$ gives 
\begin{equation*}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 \end{equation*}Using $t = 14$ gives $T =116 \left(\frac{55}{58}\right)^{\frac{14}{5}} + 55\approx 155.00^\circ$}

\end{question}

Download Question and Solution Environment\(\LaTeX\)

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\begin{document}\begin{question}(10pts) The question goes here!
    \soln{9cm}{The solution goes here.}

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HTML for Canvas

<p> <p>A coffee with temperature  <img class="equation_image" title=" \displaystyle 171^\circ " src="/equation_images/%20%5Cdisplaystyle%20171%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 171^\circ " data-equation-content=" \displaystyle 171^\circ " />  is left in a room with temperature  <img class="equation_image" title=" \displaystyle 55^\circ " src="/equation_images/%20%5Cdisplaystyle%2055%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 55^\circ " data-equation-content=" \displaystyle 55^\circ " /> . After 5 minutes the temperature of the coffee is  <img class="equation_image" title=" \displaystyle 165^\circ " src="/equation_images/%20%5Cdisplaystyle%20165%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 165^\circ " data-equation-content=" \displaystyle 165^\circ " /> , what is the temperature of the coffee after 14 minutes?</p> </p>

HTML for Canvas

<p> <p>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.  <img class="equation_image" title="  \frac{dT}{dt} = k(T(t)-T_{\text{room}}) " src="/equation_images/%20%20%5Cfrac%7BdT%7D%7Bdt%7D%20%3D%20k%28T%28t%29-T_%7B%5Ctext%7Broom%7D%7D%29%20" alt="LaTeX:   \frac{dT}{dt} = k(T(t)-T_{\text{room}}) " data-equation-content="  \frac{dT}{dt} = k(T(t)-T_{\text{room}}) " /> Using the substitution  <img class="equation_image" title=" \displaystyle y(t)=T(t)-55 " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%3DT%28t%29-55%20" alt="LaTeX:  \displaystyle y(t)=T(t)-55 " data-equation-content=" \displaystyle y(t)=T(t)-55 " />  and calculating the derivative gives  <img class="equation_image" title=" \displaystyle \frac{dy}{dt}=\frac{dT}{dt} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7Bdy%7D%7Bdt%7D%3D%5Cfrac%7BdT%7D%7Bdt%7D%20" alt="LaTeX:  \displaystyle \frac{dy}{dt}=\frac{dT}{dt} " data-equation-content=" \displaystyle \frac{dy}{dt}=\frac{dT}{dt} " /> . Calculating the new initial condition using the point  <img class="equation_image" title=" \displaystyle (5, 165) " src="/equation_images/%20%5Cdisplaystyle%20%285%2C%20165%29%20" alt="LaTeX:  \displaystyle (5, 165) " data-equation-content=" \displaystyle (5, 165) " />  and the substition gives  <img class="equation_image" title=" \displaystyle y(0) = T(0)-55 = 116 " src="/equation_images/%20%5Cdisplaystyle%20y%280%29%20%3D%20T%280%29-55%20%3D%20116%20" alt="LaTeX:  \displaystyle y(0) = T(0)-55 = 116 " data-equation-content=" \displaystyle y(0) = T(0)-55 = 116 " /> . The point  <img class="equation_image" title=" \displaystyle (5, 165) " src="/equation_images/%20%5Cdisplaystyle%20%285%2C%20165%29%20" alt="LaTeX:  \displaystyle (5, 165) " data-equation-content=" \displaystyle (5, 165) " />  must also be transformed to get  <img class="equation_image" title=" \displaystyle y(5) = T(5)-55 = 165 - 55 = 110 " src="/equation_images/%20%5Cdisplaystyle%20y%285%29%20%3D%20T%285%29-55%20%3D%20165%20-%2055%20%3D%20110%20" alt="LaTeX:  \displaystyle y(5) = T(5)-55 = 165 - 55 = 110 " data-equation-content=" \displaystyle y(5) = T(5)-55 = 165 - 55 = 110 " /> . Substituting both of these into the equation gives the new equaiton  <img class="equation_image" title=" \displaystyle \frac{dy}{dt}=ky " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7Bdy%7D%7Bdt%7D%3Dky%20" alt="LaTeX:  \displaystyle \frac{dy}{dt}=ky " data-equation-content=" \displaystyle \frac{dy}{dt}=ky " />  which has the solution  <img class="equation_image" title=" \displaystyle y(t) = y(0)e^{kt}=116e^{kt} " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%20%3D%20y%280%29e%5E%7Bkt%7D%3D116e%5E%7Bkt%7D%20" alt="LaTeX:  \displaystyle y(t) = y(0)e^{kt}=116e^{kt} " data-equation-content=" \displaystyle y(t) = y(0)e^{kt}=116e^{kt} " /> . Evaluating the function at the point gives  <img class="equation_image" title=" \displaystyle 110=116e^{5k} " src="/equation_images/%20%5Cdisplaystyle%20110%3D116e%5E%7B5k%7D%20" alt="LaTeX:  \displaystyle 110=116e^{5k} " data-equation-content=" \displaystyle 110=116e^{5k} " />  and isolating the exponential gives  <img class="equation_image" title=" \displaystyle \frac{55}{58}=e^{5k} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B55%7D%7B58%7D%3De%5E%7B5k%7D%20" alt="LaTeX:  \displaystyle \frac{55}{58}=e^{5k} " data-equation-content=" \displaystyle \frac{55}{58}=e^{5k} " /> . Solving for  <img class="equation_image" title=" \displaystyle k " src="/equation_images/%20%5Cdisplaystyle%20k%20" alt="LaTeX:  \displaystyle k " data-equation-content=" \displaystyle k " />  gives  <img class="equation_image" title=" \displaystyle k=\frac{\ln{\left(\frac{55}{58} \right)}}{5} " src="/equation_images/%20%5Cdisplaystyle%20k%3D%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B55%7D%7B58%7D%20%5Cright%29%7D%7D%7B5%7D%20" alt="LaTeX:  \displaystyle k=\frac{\ln{\left(\frac{55}{58} \right)}}{5} " data-equation-content=" \displaystyle k=\frac{\ln{\left(\frac{55}{58} \right)}}{5} " /> .  Substuting  <img class="equation_image" title=" \displaystyle k " src="/equation_images/%20%5Cdisplaystyle%20k%20" alt="LaTeX:  \displaystyle k " data-equation-content=" \displaystyle k " />  back into the equation gives  <img class="equation_image" title=" \displaystyle y(t) = 116e^{\frac{\ln{\left(\frac{55}{58} \right)}}{5}t} " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%20%3D%20116e%5E%7B%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B55%7D%7B58%7D%20%5Cright%29%7D%7D%7B5%7Dt%7D%20" alt="LaTeX:  \displaystyle y(t) = 116e^{\frac{\ln{\left(\frac{55}{58} \right)}}{5}t} " data-equation-content=" \displaystyle y(t) = 116e^{\frac{\ln{\left(\frac{55}{58} \right)}}{5}t} " />  and simplifying gives  <img class="equation_image" title=" \displaystyle y(t) = 116 \left(\frac{55}{58}\right)^{\frac{t}{5}} " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%20%3D%20116%20%5Cleft%28%5Cfrac%7B55%7D%7B58%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B5%7D%7D%20" alt="LaTeX:  \displaystyle y(t) = 116 \left(\frac{55}{58}\right)^{\frac{t}{5}} " data-equation-content=" \displaystyle y(t) = 116 \left(\frac{55}{58}\right)^{\frac{t}{5}} " /> . Substituting out  <img class="equation_image" title=" \displaystyle y(t) " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%20" alt="LaTeX:  \displaystyle y(t) " data-equation-content=" \displaystyle y(t) " />  gives 
 <img class="equation_image" title=" 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  " src="/equation_images/%20T%28t%29-55%20%3D%20116%20%5Cleft%28%5Cfrac%7B55%7D%7B58%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B5%7D%7D%20%5Cimplies%5C%2C%20T%28t%29%3D%20116%20%5Cleft%28%5Cfrac%7B55%7D%7B58%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B5%7D%7D%20%2B%2055%20%20" alt="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  " data-equation-content=" 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  <img class="equation_image" title=" \displaystyle t = 14 " src="/equation_images/%20%5Cdisplaystyle%20t%20%3D%2014%20" alt="LaTeX:  \displaystyle t = 14 " data-equation-content=" \displaystyle t = 14 " />  gives  <img class="equation_image" title=" \displaystyle T =116 \left(\frac{55}{58}\right)^{\frac{14}{5}} + 55\approx 155.00^\circ " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D116%20%5Cleft%28%5Cfrac%7B55%7D%7B58%7D%5Cright%29%5E%7B%5Cfrac%7B14%7D%7B5%7D%7D%20%2B%2055%5Capprox%20155.00%5E%5Ccirc%20" alt="LaTeX:  \displaystyle T =116 \left(\frac{55}{58}\right)^{\frac{14}{5}} + 55\approx 155.00^\circ " data-equation-content=" \displaystyle T =116 \left(\frac{55}{58}\right)^{\frac{14}{5}} + 55\approx 155.00^\circ " /> </p> </p>