Question

1. One of the realizations of English physicist Sir Isaac Newton was that

The rate of heat loss of a body is proportional to the temperature difference between the body and its surroundings.
or, if T is the temperature of the object, A the temperature of the surroundings, and k the rate of cooling

dT/dt = - k( T - A )

Suppose Anduril has just been reforged from the shards of Narsil and is placed in a basin of water to cool. The initial temperature of the sword is 1300 degrees Fahrenheit and the temperature of the water is 60 degrees Fahrenheit. If the cooling rate of the sword...

Answer 1

(when placed in water) is k = .368, find how long it takes the sword to cool to 75 degrees.
In particular you should:
_ Solve the differential equation.
_ Solve for the constant of integration using the initial condition.
_ Use the resulting equation to solve for t in terms of T to get the final answer.

Answer 2

first, start with filling dT/dt = -k(T-A) you can at least do that much, right?

I mean i can solve it for you but you wouldn't learn much from it.

Answer 3

@JonC are you online?

Answer 4

could you please explain it with details how it's solve

Answer 5

\[\Large \frac{ dT }{ dt }=-k \left( T-A \right)\]

\[\Large \frac{ dT }{ dt }=-0.368\left( T-60 \right)\]
and y(0) =1300

Answer 6

\[\Large \frac{ dT }{ dt }+0.368T = 60*0.368\]

use μ(t) as integrating factor so that we can apply the reverse chain rule to rewrite the equation.

\[\Large \color{red}{T' \times \mu} +\color{green}{0.368 T \times \mu}=22.08 \mu\]

the chain rule \[\Large \left( T(t) \times \mu (t) \right)'= \color{red} T \color{red} '\color{red}\times \color{red}\mu + \color{green} T \color{green} \times \ \color{green} \mu \color{green} ' \]

we already have the red part, so we don't need to change that. But to make sure we can integrate the red and green part combined, we need to make sure the green part of our equation matches the green part of the chain rule. If we find a value for mu that does this, we can apply the chain rule in reverse

Answer 7

so, let \[\large \color{red}T \times \mu '= 0.368 \color{red} T \times \mu\]

Since we have a T on both sides we can leave that out.

as you probably know, \[\Large u'= \frac{ d \mu }{ dt }\] it's just a different way of writing.

so let's write \[\Large \frac{ d \mu }{ dt } = 0.368 \mu\]
from which we can make \[\Large \frac{ d \mu }{ \mu } = 0.368 dt\]

by dividing both sides by mu and multiplying both sides with dt

now integrate\[\Large \int\limits_{}^{}\frac{ 1 }{ \mu }d \mu=\int\limits_{}^{}0.368dt\]

which becomes \[\Large \ln \left| \mu \right|=0.368t\]

Answer 8

\[\huge e ^{\ln \left| \mu \right|}=\mu \]
\[\Large \mu=e ^{0.368t}\]

Answer 9

now that we know the value of mu, let's insert it into the equation, knowing that by definition the left side of the equation it (T*mu)'

Answer 10

\[\Large \frac{ dT }{ dt }Te^{0.368t} =22.08e^{0.368t}\]

Answer 11

integrate both sides with respect to t\[\Large Te^{0.368t}=\int\limits_{}^{}22.08e^{0.368t}dt=60e^{0.368t}+C\]

Answer 12

divide both sides by e^0.368t to get the general solution

\[\Huge T=60+Ce^{-0.368t}\]

we can input the initial value T(0)=1600

\[\large 1600=60+Ce^{0}\]

from which you can easily see C=1540

so the solution we are looking for is

\[\huge T=60+1540e^{-0.368t}\]

This equation is the graph of Temperature related to time (Temperature on the y axis and time on the y axis). Now if we want to know at which time t the temperature T is 75 we must solve this equation

\[\Large 75=60+1540e^{-0.368t}\]

Answer 13

\[\Large15=1540e^{-0.368t}\]

\[\Large \frac{ 15 }{ 1540 }=\frac{ 3 }{ 308 }=e^{ -0.368t}\]

\[\Large \ln (\frac{ 3 }{ 308 }) = -0.368t\]
\[\Huge \frac{ \ln( \frac{ 3 }{ 308 } )}{ -0.368 }=t \approx 12.5856\]

Answer 14

∫22.08 e^0.368t dt if we integrate it
the answer will be 0.878087 e^(1.13884 t) +C
but you have 60 e^0.368t +C

Answer 15

how did you arrive at that number?

Answer 16

because i used a scientific calculator and i checked it but it's actually 60e^0.368t

Answer 17

sorry my bad you are right

Answer 18

it's ok.

Answer 19

what is mu means ?

Answer 20

Thank you

Answer 21

mu is how you pronounce this \(\mu\) greek letter