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Newton's Law of Cooling equation

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Problem
From Newton's Law of Cooling, we can use the differential equation

dT/dt= -k(T-Ts)

where Ts is the surrounding temperature, k is a positive constant, and T is the temperature.

Let [tex]\tau[/tex] be the time at which the initial temperature difference T0-Ts has been reduced by half. Find the relation between k and [tex]\tau[/tex].


Work

I have found the temperature equation for the object to be

T(t) = Ts + (T0-Ts)e-kt

If the initial temperature difference is reduced by half, i tried

T(t) = Ts + 0.5(T0-Ts)e-kt

but couldn't solve it any further. Could someone please shed some light on what i should do next?
 

Answers and Replies

  • #2
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[
I have found the temperature equation for the object to be

T(t) = Ts + (T0-Ts)e-kt
This seems to be right, but here:

If the initial temperature difference is reduced by half, i tried

T(t) = Ts + 0.5(T0-Ts)e-kt
you made some mistake with the parameter t. If we set notation

[tex]
\Delta T(t) = T(t) - T_s,
[/tex]

then the equation you want to solve is

[tex]
\Delta T(\tau) = \frac{1}{2}\Delta T(0).
[/tex]
 
  • #3
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Oh thanks for replying, i was able to get the correct answer.

k [tex]\tau[/tex] = ln2

However, I'm still a bit unclear of why T([tex]\tau[/tex])-Ts and T0-Ts canceled out during calculation. Or do T([tex]\tau[/tex]) and T0 equal? Also why isn't the 0.5 placed like

T([tex]\tau[/tex]) = Ts + 0.5(T0-Ts)e-k[tex]\tau[/tex], but instead

0.5[T([tex]\tau[/tex])-Ts] = (T0-Ts)e-k[tex]\tau[/tex]


Thanks again!
 
Last edited:
  • #4
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Oh thanks for replying, i was able to get the correct answer.

k [tex]\tau[/tex] = ln2
The use of latex like this doesn't seem to be working very well. At least my browser makes that look like [tex]k^{\tau}=\textrm{ln}(2)[/tex], although you are most certainly meaning [tex]k\tau = \textrm{ln}(2)[/tex].

However, I'm still a bit unclear of why T([tex]\tau[/tex])-Ts and T0-Ts canceled out during calculation. Or do T([tex]\tau[/tex]) and T0 equal? Also why isn't the 0.5 placed like

T([tex]\tau[/tex]) = Ts + 0.5(T0-Ts)e-k[tex]\tau[/tex], but instead

0.5[T([tex]\tau[/tex])-Ts] = (T0-Ts)e-k[tex]\tau[/tex]
It looks like you are getting the correct result, because you know what it is, and you are arriving at it "by force". :wink: Just try to chop the problem into smaller pieces, so that you can control it. We want this:

[tex]
\Delta T(\tau) = \frac{1}{2} \Delta T(0)\quad\quad\quad\quad (1)
[/tex]

where the temperature difference at each instant is given by

[tex]
\Delta T(t) = T(t) - T_s = (T_0-T_s) e^{-k t}.
[/tex]

So onto the left side of the equation (1) we get

[tex]
(T_0 - T_s)e^{-k\tau}
[/tex]

and on the right side

[tex]
\frac{1}{2} (T_0 - T_s)e^{-k\cdot 0}.
[/tex]

See how things go then?
 
  • #5
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Oh yeah i see it now, thanks a lot jostpuur. As for the latex inputs, I think I will go look for some tutorials on this forum. :rofl:
 

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