Question Statement

A thermometer showing room temperature of $80^{\circ }\mathrm{F}$ is placed on a block of ice with a temperature of $30^{\circ }\mathrm{F}$. After one minute the temperature of thermometer is $40^{\circ }\mathrm{F}$. How long will it take for the thermometer to have a temperature of $70^{\circ }\mathrm{F}$?

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Background and Explanation

This problem is solved using Newton's Law of Cooling, which states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient (surrounding) temperature. We use a first-order differential equation to model this relationship and solve for the unknown time.

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Solution

According to Newton's Law of Cooling:

$\frac{dT}{dt}=-k(T-T_s)$

Where:

• $T$ is the temperature of the thermometer at any time $t$.
• $T_s$ is the surrounding temperature (in this case, the temperature of the ice).
• $k$ is the cooling constant.

Given the surrounding temperature $T_s=30^{\circ }\mathrm{F}$:

$\begin{array}{rl}\frac{dT}{dt}& =-k(T-30)\\ \frac{dT}{T-30}& =-kdt\end{array}$

Step 1: Integrate the Differential Equation

Integrating both sides to find the general solution:

\begin{align*} \int \frac{d T}{T-30} & =-k \int d t \\ \ln (T-30) & =-k t+A \end{align*}

Step 2: Find the Constant of Integration ($A$)

We use the initial condition: at $t=0$, the thermometer is at room temperature, so $T=80$.

$\begin{array}{rl}\ln (80-30)& =-k(0)+A\\ \ln 50& =0+A\\ A& =\ln (50)\end{array}$

Substitute $A$ back into equation (1):

\begin{align*} \ln (T-30) & =-k t+\ln 50 \\ \ln (T-30)-\ln 50 & =-k t \\ \ln \left(\frac{T-30}{50}\right) & =-k t \end{align*}

Step 3: Find the Cooling Constant ($k$)

We are given that after one minute ($t=1$), the temperature $T=40$. Substitute these values into equation (2):

$\begin{array}{rl}\ln \left(\frac{40-30}{50}\right)& =-k(1)\\ \ln \left(\frac{10}{50}\right)& =-k\\ \ln \left(\frac{1}{5}\right)& =-k\\ \ln 1-\ln 5& =-k\\ 0-\ln 5& =-k\\ -\ln 5& =-k\\ k& =\ln 5\approx 1.6094\end{array}$

Substitute the value of $k$ back into equation (2):

$\ln \left(\frac{T-30}{50}\right)=-(1.6094)t$

Step 4: Calculate the time for $T=70^{\circ }\mathrm{F}$

Now we find $t$ when $T=70$. Using equation (3):

$\begin{array}{rl}\ln \left(\frac{70-30}{50}\right)& =-(1.6094)t\\ \ln \left(\frac{40}{50}\right)& =-(1.6094)t\\ \ln (0.8)& =-(1.6094)t\\ -0.2231& =-(1.6094)t\\ t& =\frac{0.2231}{1.6094}\\ t& \approx 0.1386\text{ minutes}\end{array}$

To convert this into seconds: $\begin{array}{rl}t& \approx 0.14\text{ minutes}\\ t& =(0.14\times 60)\text{ seconds}\\ t& \approx 8.3\text{ seconds}\end{array}$

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Key Formulas or Methods Used

• Newton's Law of Cooling: $\frac{dT}{dt}=-k(T-T_s)$
• Separation of Variables: A method for solving ordinary differential equations.
• Logarithmic Properties: $\ln (a)-\ln (b)=\ln (\frac{a}{b})$ and $\ln (1)=0$.

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Summary of Steps

1. Set up the differential equation using the surrounding temperature ($30^{\circ }\mathrm{F}$).
2. Separate the variables and integrate to get the logarithmic form of the equation.
3. Use the initial temperature ($80^{\circ }\mathrm{F}$ at $t=0$) to solve for the integration constant $A$.
4. Use the known temperature at $t=1$ minute to solve for the cooling constant $k$.
5. Plug the target temperature ($70^{\circ }\mathrm{F}$) into the equation and solve for the final time $t$.