Respuesta :
The result for the data given are-
Part A: The differential equation for T is [tex]\frac{d T}{d t}=\frac{1}{2} \ln \left(\frac{25}{41}\right)(T+18)[/tex].
Part B: The temperature of the chili after 4 hours is -2.744°C.
Part C: The time, t, will the chili's temperature be −10°C is 6.615 hours.
What is Newton's Law of Cooling?
According to Newton's law of cooling, the rate that a body cools directly proportional to the temperature differential in between object and its surroundings. Simply explained, a mug of hot water cools quicker in a cold environment than it does in a warm room.
Now, according to the given conditions;
Based on the question, the change rate of any object specified by T is directly related to the difference between T as well as the temperature of a surrounding environment around it, denoted by X.
[tex]\frac{d T}{d t}=k(T-X)[/tex]
Here, K is a proportionality constant. And the surrounding temperature is stated to be (-18°C). Thus,
[tex]\frac{d T}{d t}=k(T+18)[/tex]
Part A requires us to solve for k in order to derive a differential equation for T.
To answer the problem, we isolate the components and then integrate.
[tex]\begin{aligned}&\int \frac{d T}{T+18}=\int k d t \\&\ln (T+18)=k t+c\end{aligned}[/tex]
The initial temperature of even a pot with chili is 23°C, therefore at this point [tex]t=0, T_{0}=23 * C[/tex]
We get the following when we substitute 23 for T & 0 for t:
[tex]\begin{aligned}&\ln (23+18)=k(0)+c \\&\ln (41)=c\end{aligned}[/tex]
We know that the temperature for chili for 2 hours equals 7°C, so we can
[tex]t=2, T_{1}=7[/tex]
substitute t for 2 and T for 7°, and we get:
[tex]\begin{aligned}&\ln (7+18)=2 k+\ln (41) \\&\ln (25)=2 k+\ln (41)\end{aligned}[/tex]
2k Problem Solving
[tex]\begin{aligned}&2 k=\ln (25)-\ln (41) \\&2 k=\ln \left(\frac{25}{41}\right) \\&k=\frac{1}{2} \ln \left(\frac{25}{41}\right)\end{aligned}[/tex]
Substituting the obtained value of the differential equation is
[tex]\frac{d T}{d t}=\frac{1}{2} \ln \left(\frac{25}{41}\right)(T+18)[/tex].
Part B requires us to solve the mentioned differential equation to determine the temperature of a chili after four hours.
The equation gives the solution to the differential equation.
[tex]\ln (T+18)=k t+c[/tex]
By changing the values for k and c, we get:
[tex]\ln (T+18)=\frac{1}{2} \ln \left(\frac{25}{41}\right) t+\ln (41)[/tex]
Using the above relationship, the temperature (T) at any time (t) can be calculated as follows.
For, [tex]t=4, T_{2}=\phi[/tex]
[tex]\begin{aligned}&\ln \left(T_{2}+18\right)=\frac{1}{2} \ln \left(\frac{25}{41}\right) * 4+\ln (41) \\&\ln \left(T_{2}+18\right)=2 \ln \left(\frac{25}{41}\right)+\ln (41) \\&\ln \left(T_{2}+18\right)=-0.989+3.714 \\&\ln \left(T_{2}+18\right) \cong 2.725\end{aligned}[/tex]
Using the natural logarithm,
[tex]\begin{aligned}&T_{2}+18=e^{2.725}=15.256 \\&T_{2}=15.256-18 \\&T_{2}=-2.744\end{aligned}[/tex]
Thus, the temperature of the chili after 4 hours is -2.744°C.
Part C: To find component C, we must replace again to determine when the chili will be 10°C.
[tex]t=\phi, T=-10[/tex]
[tex]\begin{aligned}&\ln (-10+18)=\frac{1}{2} \ln \left(\frac{25}{41}\right) t+\ln (41) \\&\ln (8)=\frac{1}{2} \ln \left(\frac{25}{41}\right) t+\ln (41)\end{aligned}[/tex]
On further solving for [tex]\frac{1}{2} \ln \left(\frac{25}{41}\right) t[/tex]
[tex]\begin{aligned}&\frac{1}{2} \ln \left(\frac{25}{41}\right) t=\ln (8)-\ln (41) \\&\frac{1}{2} \ln \left(\frac{25}{41}\right) t=\ln \left(\frac{8}{41}\right) \\&\frac{1}{2} \ln \left(\frac{25}{41}\right) t=-1.634 \\&\ln \left(\frac{25}{41}\right) t=-1.634 * 2 \\&(-0.494) t=-3.268 \\&t=\frac{-3.268}{-0.491}\end{aligned}[/tex]
t = 6.615 hours approx.
Thus, the time, t, will the chili's temperature be −10°C is 6.615 hours.
To know more about Newton's Law of Cooling, here
https://brainly.com/question/13748261
#SPJ4
