Newton's Cooling Temperature formula problem

In summary: T(t) = Te + (T[0]+Te)e^-rtWhere r = -0.001At T(240) = 79.210=> T(240) = 79.210 = T[e] + (T[0]+Te)e^-rt
  • #1
Lami
5
0
*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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  • #2
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?
 
  • #3
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.
 
  • #4
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##
 
  • #5
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.
 
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  • #6
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.
 
  • #7
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.
 

FAQ: Newton's Cooling Temperature formula problem

What is Newton's Cooling Temperature formula?

Newton's Cooling Temperature formula, also known as Newton's Law of Cooling, is a mathematical equation used to describe the rate at which a hot object cools down to the surrounding temperature.

What is the formula for Newton's Cooling Temperature?

The formula for Newton's Cooling Temperature is T(t) = Ts + (T0 - Ts)e-kt, where T(t) represents the temperature of the object at a specific time, Ts is the surrounding temperature, T0 is the initial temperature of the object, and k is a constant representing the rate of cooling.

What are the assumptions made in Newton's Cooling Temperature formula?

There are three main assumptions made in Newton's Cooling Temperature formula: 1) the object is in a uniform environment, 2) the temperature difference between the object and the surrounding is small, and 3) the rate of heat transfer is proportional to the temperature difference between the object and the surrounding.

How is Newton's Cooling Temperature formula used in real life?

Newton's Cooling Temperature formula is commonly used in various industries, such as food and beverage, pharmaceuticals, and electronics, to determine the cooling rate of hot objects and to optimize cooling processes. It is also used in meteorology to study the cooling of the Earth's surface.

Are there any limitations to Newton's Cooling Temperature formula?

While Newton's Cooling Temperature formula is a useful tool for predicting cooling rates, it does have limitations. It assumes a constant surrounding temperature and does not take into account factors such as wind or humidity. It also does not apply to objects with non-uniform shapes or sizes, or objects with varying initial temperatures.

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