Joan Thompson

2021-12-27

A thermometer reading ${75}^{\circ }F$ is taken out where the temperature is ${20}^{\circ }F$. The reading is ${30}^{\circ }F$ 4 min later.
A. What is the value of k in four decimal places?
B. Find the thermometer reading 7 min after the thermometer was brought outside.
C. Find the time taken for the reading to drop from ${75}^{\circ }F$ to within a half degree of the air temperature.

boronganfh

Step 1
Given that:
- Initial temperature $={75}^{\circ }F$
- Surrounding temperature $={20}^{\circ }F$
- The temperature is ${30}^{\circ }F$ after 4 minute
Step 2
According to the Newtons

ambarakaq8

Solution:
$\frac{dT}{dt}=-k\left(T-{T}_{s}\right)$
$\frac{dT}{dt}=-k\left(T-20\right)$

$\mathrm{ln}\left(T-20\right)=-kt+\mathrm{ln}C$
$\mathrm{ln}\left(T-20\right)={\mathrm{ln}e}^{-kt}+\mathrm{ln}C$
$\mathrm{ln}\left(T-20\right)=\mathrm{ln}C{e}^{-kt}$
$T-20=C{e}^{-kt}$
$T=20+C{e}^{-kt}$
When $t=0,T={75}^{\circ }$
$75=20+C{e}^{0}$
$C=55$
When $t=4,T={30}^{\circ }$
$30=20+55{e}^{-4k}$
$\frac{10}{55}={e}^{-4k}$
${e}^{-k}={\left(\frac{2}{11}\right)}^{\frac{1}{4}}$
Thus, $T=20+55{\left(\frac{2}{11}\right)}^{\frac{t}{4}}$
When $T={20.5}^{\circ }F$
$20.5=20+55{\left(\frac{2}{11}\right)}^{\frac{t}{4}}$
$\frac{0.5}{55}={\left(\frac{2}{11}\right)}^{\frac{t}{4}}$
$\mathrm{ln}\frac{0.5}{55}={\mathrm{ln}\left(\frac{2}{11}\right)}^{\frac{t}{4}}$
$\mathrm{ln}\left(\frac{1}{110}\right)=\frac{1}{4}t\mathrm{ln}\left(\frac{2}{11}\right)$
$t=\frac{4\mathrm{ln}\left(\frac{1}{110}\right)}{\mathrm{ln}\left(\frac{2}{11}\right)}$

Vasquez

user_27qwe

Result:
$t\approx 12.3413$ minutes
Solution:
A. To solve for the value of k, we can use the formula for exponential decay:
$T\left(t\right)={T}_{0}{e}^{-kt}$,
where $T\left(t\right)$ is the temperature at time t, ${T}_{0}$ is the initial temperature, and k is the decay constant.
Given that the initial temperature ${T}_{0}$ is 75 degrees F and the temperature at time t = 4 minutes is 30 degrees F, we have:
$30=75{e}^{-4k}$.
To solve for k, we can divide both sides by 75 and take the natural logarithm of both sides:
$\mathrm{ln}\left(\frac{30}{75}\right)=-4k$.
Simplifying further:
$\mathrm{ln}\left(\frac{2}{5}\right)=-4k$.
Finally, we can solve for k by dividing both sides by -4:
$k=-\frac{\mathrm{ln}\left(\frac{2}{5}\right)}{4}$.
Thus, the value of k, rounded to four decimal places, is approximately $k\approx 0.2835$.
B. To find the thermometer reading 7 minutes after it was brought outside, we can use the same formula as before:
$T\left(t\right)={T}_{0}{e}^{-kt}$.
Given that the initial temperature ${T}_{0}$ is 75 degrees F and t = 7 minutes, we can substitute these values into the formula:
$T\left(7\right)=75{e}^{-0.2835·7}$.
Evaluating this expression, we find:
$T\left(7\right)=75{e}^{-1.9845}$.
Thus, the thermometer reading 7 minutes after it was brought outside is approximately $T\left(7\right)\approx 27.2571$ degrees F.
C. To find the time taken for the reading to drop from 75 degrees F to within a half degree of the air temperature, we need to determine the time it takes for the temperature to reach a value of 75 - 0.5 = 74.5 degrees F.
Using the formula $T\left(t\right)={T}_{0}{e}^{-kt}$ and substituting ${T}_{0}=75$ and $T\left(t\right)=74.5$, we have:
$74.5=75{e}^{-kt}$.
Dividing both sides by 75 and taking the natural logarithm of both sides, we get:
$\mathrm{ln}\left(\frac{74.5}{75}\right)=-kt$.
Simplifying further:
$\mathrm{ln}\left(\frac{298}{300}\right)=-kt$.
Finally, we can solve for t by dividing both sides by -k:
$t=\frac{\mathrm{ln}\left(\frac{300}{298}\right)}{k}$.
Using the previously calculated value of k, we can substitute it into the equation and evaluate t:
$t=\frac{\mathrm{ln}\left(\frac{300}{298}\right)}{0.2835}$.
Thus, the time taken for the reading to drop from 75 degrees F to within a half degree of the air temperature is approximately $t\approx 12.3413$ minutes.

karton

A. To find the value of k, we can use the exponential decay formula:
$T\left(t\right)={T}_{0}+\left({T}_{a}-{T}_{0}\right)·{e}^{-kt}$,
where:
- $T\left(t\right)$ is the temperature at time t,
- ${T}_{0}$ is the initial temperature of the thermometer (75°F),
- ${T}_{a}$ is the ambient temperature (20°F),
- $k$ is the decay constant, and
- $t$ is the time elapsed.
We are given that the temperature dropped to 30°F after 4 minutes. Plugging in the values:
$30=75+\left(20-75\right)·{e}^{-k·4}$.
To solve for k, we can rearrange the equation as follows:
$\frac{30-75}{20-75}={e}^{-4k}$.
Taking the natural logarithm of both sides, we get:
$\mathrm{ln}\left(\frac{30-75}{20-75}\right)=-4k$.
Now we can solve for k:
$k=\frac{\mathrm{ln}\left(\frac{30-75}{20-75}\right)}{-4}$.
Calculating the value of k using a calculator or computer program, we find:
$k\approx -0.0784$ (rounded to four decimal places).
Therefore, the value of k is approximately $-0.0784$.
B. To find the thermometer reading 7 minutes after it was brought outside, we can use the same exponential decay formula:
$T\left(t\right)={T}_{0}+\left({T}_{a}-{T}_{0}\right)·{e}^{-kt}$.
Plugging in the values:
$T\left(7\right)=75+\left(20-75\right)·{e}^{-0.0784·7}$.
Calculating this using a calculator or computer program, we find:
$T\left(7\right)\approx 35.5073$ (rounded to four decimal places).
Therefore, the thermometer reading 7 minutes after it was brought outside is approximately $35.5073$°F.
C. To find the time taken for the reading to drop from 75°F to within a half degree of the air temperature, we need to solve the following equation:
$T\left(t\right)={T}_{a}+\frac{{T}_{0}-{T}_{a}}{2}$.
Plugging in the values:
$T\left(t\right)=20+\frac{75-20}{2}$.
Simplifying the equation:
$T\left(t\right)=47.5$.
Now we can solve for t:
$47.5=75+\left(20-75\right)·{e}^{-0.0784·t}$.
Rearranging the equation:
${e}^{-0.0784·t}=\frac{47.5-75}{20-75}$.
Taking the natural logarithm of both sides:
$-0.0784·t=\mathrm{ln}\left(\frac{47.5-75}{20-75}\right)$.
Solving for t:
$t=\frac{\mathrm{ln}\left(\frac{47.5-75}{20-75}\right)}{-0.0784}$.
Calculating the value of t using a calculator or computer program, we find:
$t\approx 8.5614$ (rounded to four decimal places).
Therefore, the time taken for the reading to drop from 75°F to within a half degree of the air temperature is approximately $8$

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