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A coffee with temperature \(\displaystyle 154^\circ\) is left in a room with temperature \(\displaystyle 59^\circ\). After 7 minutes the temperature of the coffee is \(\displaystyle 149^\circ\), how long until the coffee is \(\displaystyle 124^\circ\)?
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. \begin{equation*} \frac{dT}{dt} = k(T(t)-T_{\text{room}})\end{equation*}Using the substitution \(\displaystyle y(t)=T(t)-59\) and calculating the derivative gives \(\displaystyle \frac{dy}{dt}=\frac{dT}{dt}\). Calculating the new initial condition using the point \(\displaystyle (7, 149)\) and the substition gives \(\displaystyle y(0) = T(0)-59 = 95\). The point \(\displaystyle (7, 149)\) must also be transformed to get \(\displaystyle y(7) = T(7)-59 = 149 - 59 = 90\). Substituting both of these into the equation gives the new equaiton \(\displaystyle \frac{dy}{dt}=ky\) which has the solution \(\displaystyle y(t) = y(0)e^{kt}=95e^{kt}\). Evaluating the function at the point gives \(\displaystyle 90=95e^{7k}\) and isolating the exponential gives \(\displaystyle \frac{18}{19}=e^{7k}\). Solving for \(\displaystyle k\) gives \(\displaystyle k=\frac{\ln{\left(\frac{18}{19} \right)}}{7}\). Substuting \(\displaystyle k\) back into the equation gives \(\displaystyle y(t) = 95e^{\frac{\ln{\left(\frac{18}{19} \right)}}{7}t}\) and simplifying gives \(\displaystyle y(t) = 95 \left(\frac{18}{19}\right)^{\frac{t}{7}}\). Substituting out \(\displaystyle y(t)\) gives \begin{equation*}T(t)-59 = 95 \left(\frac{18}{19}\right)^{\frac{t}{7}} \implies\, T(t)= 95 \left(\frac{18}{19}\right)^{\frac{t}{7}} + 59 \end{equation*}Using \(\displaystyle T\) gives the equation \(\displaystyle 124=95 \left(\frac{18}{19}\right)^{\frac{t}{7}} + 59\). Isolating the exponential gives \(\displaystyle \frac{13}{19}=\left(\frac{18}{19}\right)^{\frac{t}{7}}\). Taking the natural logarithm of both sides and solving for \(\displaystyle t\) gives \(\displaystyle t = \frac{7 \ln{\left(\frac{13}{19} \right)}}{\ln{\left(\frac{18}{19} \right)}}\approx 49.1\) minutes.
\begin{question}A coffee with temperature $154^\circ$ is left in a room with temperature $59^\circ$. After 7 minutes the temperature of the coffee is $149^\circ$, how long until the coffee is $124^\circ$?
\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 $y(t)=T(t)-59$ and calculating the derivative gives $\frac{dy}{dt}=\frac{dT}{dt}$. Calculating the new initial condition using the point $(7, 149)$ and the substition gives $y(0) = T(0)-59 = 95$. The point $(7, 149)$ must also be transformed to get $y(7) = T(7)-59 = 149 - 59 = 90$. Substituting both of these into the equation gives the new equaiton $\frac{dy}{dt}=ky$ which has the solution $y(t) = y(0)e^{kt}=95e^{kt}$. Evaluating the function at the point gives $90=95e^{7k}$ and isolating the exponential gives $\frac{18}{19}=e^{7k}$. Solving for $k$ gives $k=\frac{\ln{\left(\frac{18}{19} \right)}}{7}$. Substuting $k$ back into the equation gives $y(t) = 95e^{\frac{\ln{\left(\frac{18}{19} \right)}}{7}t}$ and simplifying gives $y(t) = 95 \left(\frac{18}{19}\right)^{\frac{t}{7}}$. Substituting out $y(t)$ gives
\begin{equation*}T(t)-59 = 95 \left(\frac{18}{19}\right)^{\frac{t}{7}} \implies\, T(t)= 95 \left(\frac{18}{19}\right)^{\frac{t}{7}} + 59 \end{equation*}Using $T$ gives the equation $124=95 \left(\frac{18}{19}\right)^{\frac{t}{7}} + 59$. Isolating the exponential gives $\frac{13}{19}=\left(\frac{18}{19}\right)^{\frac{t}{7}}$. Taking the natural logarithm of both sides and solving for $t$ gives $t = \frac{7 \ln{\left(\frac{13}{19} \right)}}{\ln{\left(\frac{18}{19} \right)}}\approx 49.1$ minutes. }
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>A coffee with temperature <img class="equation_image" title=" \displaystyle 154^\circ " src="/equation_images/%20%5Cdisplaystyle%20154%5E%5Ccirc%20" alt="LaTeX: \displaystyle 154^\circ " data-equation-content=" \displaystyle 154^\circ " /> is left in a room with temperature <img class="equation_image" title=" \displaystyle 59^\circ " src="/equation_images/%20%5Cdisplaystyle%2059%5E%5Ccirc%20" alt="LaTeX: \displaystyle 59^\circ " data-equation-content=" \displaystyle 59^\circ " /> . After 7 minutes the temperature of the coffee is <img class="equation_image" title=" \displaystyle 149^\circ " src="/equation_images/%20%5Cdisplaystyle%20149%5E%5Ccirc%20" alt="LaTeX: \displaystyle 149^\circ " data-equation-content=" \displaystyle 149^\circ " /> , how long until the coffee is <img class="equation_image" title=" \displaystyle 124^\circ " src="/equation_images/%20%5Cdisplaystyle%20124%5E%5Ccirc%20" alt="LaTeX: \displaystyle 124^\circ " data-equation-content=" \displaystyle 124^\circ " /> ?</p> </p>
<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)-59 " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%3DT%28t%29-59%20" alt="LaTeX: \displaystyle y(t)=T(t)-59 " data-equation-content=" \displaystyle y(t)=T(t)-59 " /> 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 (7, 149) " src="/equation_images/%20%5Cdisplaystyle%20%287%2C%20149%29%20" alt="LaTeX: \displaystyle (7, 149) " data-equation-content=" \displaystyle (7, 149) " /> and the substition gives <img class="equation_image" title=" \displaystyle y(0) = T(0)-59 = 95 " src="/equation_images/%20%5Cdisplaystyle%20y%280%29%20%3D%20T%280%29-59%20%3D%2095%20" alt="LaTeX: \displaystyle y(0) = T(0)-59 = 95 " data-equation-content=" \displaystyle y(0) = T(0)-59 = 95 " /> . The point <img class="equation_image" title=" \displaystyle (7, 149) " src="/equation_images/%20%5Cdisplaystyle%20%287%2C%20149%29%20" alt="LaTeX: \displaystyle (7, 149) " data-equation-content=" \displaystyle (7, 149) " /> must also be transformed to get <img class="equation_image" title=" \displaystyle y(7) = T(7)-59 = 149 - 59 = 90 " src="/equation_images/%20%5Cdisplaystyle%20y%287%29%20%3D%20T%287%29-59%20%3D%20149%20-%2059%20%3D%2090%20" alt="LaTeX: \displaystyle y(7) = T(7)-59 = 149 - 59 = 90 " data-equation-content=" \displaystyle y(7) = T(7)-59 = 149 - 59 = 90 " /> . 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}=95e^{kt} " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%20%3D%20y%280%29e%5E%7Bkt%7D%3D95e%5E%7Bkt%7D%20" alt="LaTeX: \displaystyle y(t) = y(0)e^{kt}=95e^{kt} " data-equation-content=" \displaystyle y(t) = y(0)e^{kt}=95e^{kt} " /> . Evaluating the function at the point gives <img class="equation_image" title=" \displaystyle 90=95e^{7k} " src="/equation_images/%20%5Cdisplaystyle%2090%3D95e%5E%7B7k%7D%20" alt="LaTeX: \displaystyle 90=95e^{7k} " data-equation-content=" \displaystyle 90=95e^{7k} " /> and isolating the exponential gives <img class="equation_image" title=" \displaystyle \frac{18}{19}=e^{7k} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B18%7D%7B19%7D%3De%5E%7B7k%7D%20" alt="LaTeX: \displaystyle \frac{18}{19}=e^{7k} " data-equation-content=" \displaystyle \frac{18}{19}=e^{7k} " /> . 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{18}{19} \right)}}{7} " src="/equation_images/%20%5Cdisplaystyle%20k%3D%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B18%7D%7B19%7D%20%5Cright%29%7D%7D%7B7%7D%20" alt="LaTeX: \displaystyle k=\frac{\ln{\left(\frac{18}{19} \right)}}{7} " data-equation-content=" \displaystyle k=\frac{\ln{\left(\frac{18}{19} \right)}}{7} " /> . 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) = 95e^{\frac{\ln{\left(\frac{18}{19} \right)}}{7}t} " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%20%3D%2095e%5E%7B%5Cfrac%7B%5Cln%7B%5Cleft%28%5Cfrac%7B18%7D%7B19%7D%20%5Cright%29%7D%7D%7B7%7Dt%7D%20" alt="LaTeX: \displaystyle y(t) = 95e^{\frac{\ln{\left(\frac{18}{19} \right)}}{7}t} " data-equation-content=" \displaystyle y(t) = 95e^{\frac{\ln{\left(\frac{18}{19} \right)}}{7}t} " /> and simplifying gives <img class="equation_image" title=" \displaystyle y(t) = 95 \left(\frac{18}{19}\right)^{\frac{t}{7}} " src="/equation_images/%20%5Cdisplaystyle%20y%28t%29%20%3D%2095%20%5Cleft%28%5Cfrac%7B18%7D%7B19%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B7%7D%7D%20" alt="LaTeX: \displaystyle y(t) = 95 \left(\frac{18}{19}\right)^{\frac{t}{7}} " data-equation-content=" \displaystyle y(t) = 95 \left(\frac{18}{19}\right)^{\frac{t}{7}} " /> . 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)-59 = 95 \left(\frac{18}{19}\right)^{\frac{t}{7}} \implies\, T(t)= 95 \left(\frac{18}{19}\right)^{\frac{t}{7}} + 59 " src="/equation_images/%20T%28t%29-59%20%3D%2095%20%5Cleft%28%5Cfrac%7B18%7D%7B19%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B7%7D%7D%20%5Cimplies%5C%2C%20T%28t%29%3D%2095%20%5Cleft%28%5Cfrac%7B18%7D%7B19%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B7%7D%7D%20%2B%2059%20%20" alt="LaTeX: T(t)-59 = 95 \left(\frac{18}{19}\right)^{\frac{t}{7}} \implies\, T(t)= 95 \left(\frac{18}{19}\right)^{\frac{t}{7}} + 59 " data-equation-content=" T(t)-59 = 95 \left(\frac{18}{19}\right)^{\frac{t}{7}} \implies\, T(t)= 95 \left(\frac{18}{19}\right)^{\frac{t}{7}} + 59 " /> Using <img class="equation_image" title=" \displaystyle T " src="/equation_images/%20%5Cdisplaystyle%20T%20" alt="LaTeX: \displaystyle T " data-equation-content=" \displaystyle T " /> gives the equation <img class="equation_image" title=" \displaystyle 124=95 \left(\frac{18}{19}\right)^{\frac{t}{7}} + 59 " src="/equation_images/%20%5Cdisplaystyle%20124%3D95%20%5Cleft%28%5Cfrac%7B18%7D%7B19%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B7%7D%7D%20%2B%2059%20" alt="LaTeX: \displaystyle 124=95 \left(\frac{18}{19}\right)^{\frac{t}{7}} + 59 " data-equation-content=" \displaystyle 124=95 \left(\frac{18}{19}\right)^{\frac{t}{7}} + 59 " /> . Isolating the exponential gives <img class="equation_image" title=" \displaystyle \frac{13}{19}=\left(\frac{18}{19}\right)^{\frac{t}{7}} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B13%7D%7B19%7D%3D%5Cleft%28%5Cfrac%7B18%7D%7B19%7D%5Cright%29%5E%7B%5Cfrac%7Bt%7D%7B7%7D%7D%20" alt="LaTeX: \displaystyle \frac{13}{19}=\left(\frac{18}{19}\right)^{\frac{t}{7}} " data-equation-content=" \displaystyle \frac{13}{19}=\left(\frac{18}{19}\right)^{\frac{t}{7}} " /> . Taking the natural logarithm of both sides and solving for <img class="equation_image" title=" \displaystyle t " src="/equation_images/%20%5Cdisplaystyle%20t%20" alt="LaTeX: \displaystyle t " data-equation-content=" \displaystyle t " /> gives <img class="equation_image" title=" \displaystyle t = \frac{7 \ln{\left(\frac{13}{19} \right)}}{\ln{\left(\frac{18}{19} \right)}}\approx 49.1 " src="/equation_images/%20%5Cdisplaystyle%20t%20%3D%20%5Cfrac%7B7%20%5Cln%7B%5Cleft%28%5Cfrac%7B13%7D%7B19%7D%20%5Cright%29%7D%7D%7B%5Cln%7B%5Cleft%28%5Cfrac%7B18%7D%7B19%7D%20%5Cright%29%7D%7D%5Capprox%2049.1%20" alt="LaTeX: \displaystyle t = \frac{7 \ln{\left(\frac{13}{19} \right)}}{\ln{\left(\frac{18}{19} \right)}}\approx 49.1 " data-equation-content=" \displaystyle t = \frac{7 \ln{\left(\frac{13}{19} \right)}}{\ln{\left(\frac{18}{19} \right)}}\approx 49.1 " /> minutes. </p> </p>