Cup of hot tea, and I need it just *warm*

  • #1

Main Question or Discussion Point

Hello,
Say I don't like to drink hot tea, in fact I don't like to drink anything hot. It is beyond my comprehension why anyone would want to drink something above 40C ;)

I have cup of hot tea 100C and i want to know how long will it take to cool down to 40C.

[tex] \frac{\Delta Q}{\Delta t}[/tex] = -k A [tex]\frac{\Delta T}{\Delta x} [/tex]

(for some reason I can't get the equation to display correctly)
so i have k of course, "delta x" would be the thickness of the mug wall, A is also given . The "delta T" part is the difference between tea and air temperature, i guess?
Now my approach would be to write an iterative algorithm, that will calculate the "instantaneous" heat flux, then lower the temperature a bit and advance time, then calculate heat flux again. Unless there is some other equation which will let me calculate it in a non-iterative way?
It seems silly that I have to iterate just to get an answer to such simple question...
This is of course if I neglect the evaporation effect, say the mug is closed, it is a fancy mug.
 

Answers and Replies

  • #2
649
2
In the limit of infinitesimal time steps time, your equation can probably be rewritten as a differential equation for the temperature like this:

[tex] \frac{dT}{dt}(t) = -b[T(t) - T_a][/tex]

where [tex]b[/tex] is some constant related to your [tex]a[/tex], [tex]k[/tex] and [tex]\Delta X[/tex], and [tex]T_a[/tex] is the ambient air temperature around the cup (assumed constant), you can integrate it to get

[tex] T(t) = [T(0) - T_a] \exp(-bt) + T_a [/tex]

Which tells your the temperature at any time. [tex]T(0)[/tex] is the tea temperature at time 0.
 

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