Newton's Cooling Temperature formula problem

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Homework Help Overview

The discussion revolves around applying Newton's Cooling formula to predict the environmental temperature (Te) based on given temperature readings at specific times. The formula involves variables such as the initial temperature (T[0]), the temperature at time t (T(t)), and the cooling rate (r).

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss substituting known values into the cooling formula and express confusion about the correct application of the formula. There are questions about how to handle the variables and whether the cooling rate should be positive or negative.

Discussion Status

Some participants have provided guidance on how to set up equations based on the temperature readings at different times. There is an acknowledgment of the need to create a second equation using the second temperature data point, indicating a productive direction in the discussion.

Contextual Notes

There is a noted confusion regarding the sign of the cooling rate (r) and its implications for the formula. Participants also mention feeling overwhelmed by the information presented, which may affect their confidence in proceeding with the problem.

Lami
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*Given Newton's Cooling formula, where T(t) predicts temperature, and where Te is the temperature of the enviornment, T[0] is the temperature when t = 0:

T(t) = Te + (T[0]+Te)e^-rt

predict Te when

T(120)= 86.632
T(240) = 79.210

r = 0.001

I don't know how to even remotely go about solving this hence the lack of the template. I feel like I'm being overloaded with information. What should be my first steps towards rationalising this?
 

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Lami said:
View attachment 231340 I don't know how to even remotely go about solving this hence the lack of the template. I feel like I'm being overloaded with information. What should be my first steps towards rationalising this?
You are given sufficient information to solve this problem. What would you get if you substituted the first data point into the equation?
 
Chestermiller said:
You are given sufficient information to solve this problem. What would you get if you substituted the first data point into the equation?

T(t) = T[e] + e^-rt (T(0) - T[e])
Where r = -0.001
At T(120) = 86.632
=> 86.632 = T[e] + e^-(-0.001*120) (T[0] - T[e])
86.632 = T[e] + T[0] e^0.12 - T[e]e^0.12

This is why I don't understand what to do because I have no idea how I should even be substituting, or what I should be substituting, into these equations after this point or if I've even went about the right way of doing it.
 
Lami said:
*Given Newton's Cooling formula, where T(t) predicts temperature, and where Te is the temperature of the environment, T[0] is the temperature when t = 0:

T(t) = Te + (T[0]+Te)e^-rt

predict Te when

T(120)= 86.632
T(240) = 79.210

r = -0.001

I don't know how to even remotely go about solving this hence the lack of the template. I feel like I'm being overloaded with information. What should be my first steps towards rationalising this?
The edited version of Post #1 is much easier to read than the image was. However, there appears to be one item that needs correction. The cooling rate, r, is a positive number. The negative sign is carried in the overall formula.

That should be: ##\ r=0.001##
 
Lami said:
T(t) = T[e] + e^-rt (T(0) - T[e])
Where r = -0.001
At T(120) = 86.632
=> 86.632 = T[e] + e^-(-0.001*120) (T[0] - T[e])
86.632 = T[e] + T[0] e^0.12 - T[e]e^0.12

This is why I don't understand what to do because I have no idea how I should even be substituting, or what I should be substituting, into these equations after this point or if I've even went about the right way of doing it.

As Sammy points out, using ##e^{-rt}## and ##r = -0.001## is wrong; you want ##e^{-0.001 t}##, so that the temperature will go from ##T_0## towards ##T_e## as time goes on. If you had ##e^{+0.001 t}## the two temperatures would grow farther apart!

Anyway, you have one almost-correct equation containing ##T_e## and ##T_0## obtained from information at ##t=120.## Write a similar equation using information at ##t = 240.## You will then have two equations in two unknowns. Quantities like e^0.12 or e^(-0.12) are just numbers that you calculate before proceeding.
 
Last edited:
Ray Vickson said:
As Sammy points out, using ##e^{-rt}## and ##r = -0.001## is wrong; that would give you ##e^{.001 t}##, so the temperature would increase to infinity as time goes on!

Anyway, you have one equation containing ##T_e## and ##T_0## obtained from information at ##t=120.## Write a similar equation using information at ##t = 240.##

I was thinking of doing that but I wasn't confident whether I could cancel but I'm now regretting my lack of confidence. Thanks.
 
Lami said:
T(t) = T[e] + e^-rt (T(0) - T[e])
Where r = -0.001
At T(120) = 86.632
=> 86.632 = T[e] + e^-(-0.001*120) (T[0] - T[e]) **
86.632 = T[e] + T[0] e^0.12 - T[e]e^0.12

This is why I don't understand what to do because I have no idea how I should even be substituting, or what I should be substituting, into these equations after this point or if I've even went about the right way of doing it.
As has been pointed out, and you acknowledged, r = (+)0.001.

So, the above equation I marked as (**) becomes:

86.632 = T[e] + e^(−0.001*120) (T[0] − T[e])​

I suggest keeping (T[0] − T[e]) together for the time being. After all, it's just some constant, the same for all values of time, t .
So that equation then becomes:

86.632 = T[e] + (T[0] − T[e]) e^(−0.120)​

As @Ray Vickson says, write the similar equation for time, t = 240 seconds.
 

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