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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 \(\displaystyle (12, 167)\) and the substition gives \(\displaystyle y(0) = T(0)-55 = 121\). The point \(\displaystyle (12, 167)\) must also be transformed to get \(\displaystyle y(12) = T(12)-55 = 167 - 55 = 112\). Substituting both of these into the equation gives the new equaiton \(\displaystyle \frac{dy}{dt}=ky\) which has the solution \(\displaystyle y(t) = y(0)e^{kt}=121e^{kt}\). Evaluating the function at the point gives \(\displaystyle 112=121e^{12k}\) and isolating the exponential gives \(\displaystyle \frac{112}{121}=e^{12k}\). Solving for \(\displaystyle k\) gives \(\displaystyle k=\frac{\ln{\left(\frac{112}{121} \right)}}{12}\). Substuting \(\displaystyle k\) back into the equation gives \(\displaystyle y(t) = 121e^{\frac{\ln{\left(\frac{112}{121} \right)}}{12}t}\) and simplifying gives \(\displaystyle y(t) = 121 \left(\frac{112}{121}\right)^{\frac{t}{12}}\). Substituting out \(\displaystyle y(t)\) gives \begin{equation*}T(t)-55 = 121 \left(\frac{112}{121}\right)^{\frac{t}{12}} \implies\, T(t)= 121 \left(\frac{112}{121}\right)^{\frac{t}{12}} + 55 \end{equation*}Using \(\displaystyle t = 24\) gives \(\displaystyle T =121 \left(\frac{112}{121}\right)^{\frac{24}{12}} + 55\approx 159.00^\circ\)

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\begin{question}A coffee with temperature $176^\circ$ is left in a room with temperature $55^\circ$. After 12 minutes the temperature of the coffee is $167^\circ$, what is the temperature of the coffee after 24 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 $(12, 167)$ and the substition gives $y(0) = T(0)-55 = 121$. The point $(12, 167)$ must also be transformed to get $y(12) = T(12)-55 = 167 - 55 = 112$. Substituting both of these into the equation gives the new equaiton $\frac{dy}{dt}=ky$ which has the solution $y(t) = y(0)e^{kt}=121e^{kt}$. Evaluating the function at the point gives $112=121e^{12k}$ and isolating the exponential gives $\frac{112}{121}=e^{12k}$. Solving for $k$ gives $k=\frac{\ln{\left(\frac{112}{121} \right)}}{12}$.  Substuting $k$ back into the equation gives $y(t) = 121e^{\frac{\ln{\left(\frac{112}{121} \right)}}{12}t}$ and simplifying gives $y(t) = 121 \left(\frac{112}{121}\right)^{\frac{t}{12}}$. Substituting out $y(t)$ gives 
\begin{equation*}T(t)-55 = 121 \left(\frac{112}{121}\right)^{\frac{t}{12}} \implies\, T(t)= 121 \left(\frac{112}{121}\right)^{\frac{t}{12}} + 55 \end{equation*}Using $t = 24$ gives $T =121 \left(\frac{112}{121}\right)^{\frac{24}{12}} + 55\approx 159.00^\circ$}

\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 176^\circ " src="/equation_images/%20%5Cdisplaystyle%20176%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 176^\circ " data-equation-content=" \displaystyle 176^\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 12 minutes the temperature of the coffee is  <img class="equation_image" title=" \displaystyle 167^\circ " src="/equation_images/%20%5Cdisplaystyle%20167%5E%5Ccirc%20" alt="LaTeX:  \displaystyle 167^\circ " data-equation-content=" \displaystyle 167^\circ " /> , what is the temperature of the coffee after 24 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 (12, 167) " src="/equation_images/%20%5Cdisplaystyle%20%2812%2C%20167%29%20" alt="LaTeX:  \displaystyle (12, 167) " data-equation-content=" \displaystyle (12, 167) " />  and the substition gives  <img class="equation_image" title=" \displaystyle y(0) = T(0)-55 = 121 " src="/equation_images/%20%5Cdisplaystyle%20y%280%29%20%3D%20T%280%29-55%20%3D%20121%20" alt="LaTeX:  \displaystyle y(0) = T(0)-55 = 121 " data-equation-content=" \displaystyle y(0) = T(0)-55 = 121 " /> . The point  <img class="equation_image" title=" \displaystyle (12, 167) " src="/equation_images/%20%5Cdisplaystyle%20%2812%2C%20167%29%20" alt="LaTeX:  \displaystyle (12, 167) " data-equation-content=" \displaystyle (12, 167) " />  must also be transformed to get  <img class="equation_image" title=" \displaystyle y(12) = T(12)-55 = 167 - 55 = 112 " src="/equation_images/%20%5Cdisplaystyle%20y%2812%29%20%3D%20T%2812%29-55%20%3D%20167%20-%2055%20%3D%20112%20" alt="LaTeX:  \displaystyle y(12) = T(12)-55 = 167 - 55 = 112 " data-equation-content=" \displaystyle y(12) = T(12)-55 = 167 - 55 = 112 " /> . 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}=121e^{kt} " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%20%3D%20y%280%29e%5E%7Bkt%7D%3D121e%5E%7Bkt%7D%20" alt="LaTeX:  \displaystyle y(t) = y(0)e^{kt}=121e^{kt} " data-equation-content=" \displaystyle y(t) = y(0)e^{kt}=121e^{kt} " /> . Evaluating the function at the point gives  <img class="equation_image" title=" \displaystyle 112=121e^{12k} " src="/equation_images/%20%5Cdisplaystyle%20112%3D121e%5E%7B12k%7D%20" alt="LaTeX:  \displaystyle 112=121e^{12k} " data-equation-content=" \displaystyle 112=121e^{12k} " />  and isolating the exponential gives  <img class="equation_image" title=" \displaystyle \frac{112}{121}=e^{12k} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B112%7D%7B121%7D%3De%5E%7B12k%7D%20" alt="LaTeX:  \displaystyle \frac{112}{121}=e^{12k} " data-equation-content=" \displaystyle \frac{112}{121}=e^{12k} " /> . 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{112}{121} \right)}}{12} " src="/equation_images/%20%5Cdisplaystyle%20k%3D%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B112%7D%7B121%7D%20%5Cright%29%7D%7D%7B12%7D%20" alt="LaTeX:  \displaystyle k=\frac{\ln{\left(\frac{112}{121} \right)}}{12} " data-equation-content=" \displaystyle k=\frac{\ln{\left(\frac{112}{121} \right)}}{12} " /> .  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) = 121e^{\frac{\ln{\left(\frac{112}{121} \right)}}{12}t} " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%20%3D%20121e%5E%7B%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B112%7D%7B121%7D%20%5Cright%29%7D%7D%7B12%7Dt%7D%20" alt="LaTeX:  \displaystyle y(t) = 121e^{\frac{\ln{\left(\frac{112}{121} \right)}}{12}t} " data-equation-content=" \displaystyle y(t) = 121e^{\frac{\ln{\left(\frac{112}{121} \right)}}{12}t} " />  and simplifying gives  <img class="equation_image" title=" \displaystyle y(t) = 121 \left(\frac{112}{121}\right)^{\frac{t}{12}} " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%20%3D%20121%20%5Cleft%28%5Cfrac%7B112%7D%7B121%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B12%7D%7D%20" alt="LaTeX:  \displaystyle y(t) = 121 \left(\frac{112}{121}\right)^{\frac{t}{12}} " data-equation-content=" \displaystyle y(t) = 121 \left(\frac{112}{121}\right)^{\frac{t}{12}} " /> . 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 = 121 \left(\frac{112}{121}\right)^{\frac{t}{12}} \implies\, T(t)= 121 \left(\frac{112}{121}\right)^{\frac{t}{12}} + 55  " src="/equation_images/%20T%28t%29-55%20%3D%20121%20%5Cleft%28%5Cfrac%7B112%7D%7B121%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B12%7D%7D%20%5Cimplies%5C%2C%20T%28t%29%3D%20121%20%5Cleft%28%5Cfrac%7B112%7D%7B121%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B12%7D%7D%20%2B%2055%20%20" alt="LaTeX:  T(t)-55 = 121 \left(\frac{112}{121}\right)^{\frac{t}{12}} \implies\, T(t)= 121 \left(\frac{112}{121}\right)^{\frac{t}{12}} + 55  " data-equation-content=" T(t)-55 = 121 \left(\frac{112}{121}\right)^{\frac{t}{12}} \implies\, T(t)= 121 \left(\frac{112}{121}\right)^{\frac{t}{12}} + 55  " /> Using  <img class="equation_image" title=" \displaystyle t = 24 " src="/equation_images/%20%5Cdisplaystyle%20t%20%3D%2024%20" alt="LaTeX:  \displaystyle t = 24 " data-equation-content=" \displaystyle t = 24 " />  gives  <img class="equation_image" title=" \displaystyle T =121 \left(\frac{112}{121}\right)^{\frac{24}{12}} + 55\approx 159.00^\circ " src="/equation_images/%20%5Cdisplaystyle%20T%20%3D121%20%5Cleft%28%5Cfrac%7B112%7D%7B121%7D%5Cright%29%5E%7B%5Cfrac%7B24%7D%7B12%7D%7D%20%2B%2055%5Capprox%20159.00%5E%5Ccirc%20" alt="LaTeX:  \displaystyle T =121 \left(\frac{112}{121}\right)^{\frac{24}{12}} + 55\approx 159.00^\circ " data-equation-content=" \displaystyle T =121 \left(\frac{112}{121}\right)^{\frac{24}{12}} + 55\approx 159.00^\circ " /> </p> </p>