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

## 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.

$$\frac{\Delta Q}{\Delta t}$$ = -k A $$\frac{\Delta T}{\Delta x}$$

(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.

Related Other Physics Topics News on Phys.org
In the limit of infinitesimal time steps time, your equation can probably be rewritten as a differential equation for the temperature like this:

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

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

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

Which tells your the temperature at any time. $$T(0)$$ is the tea temperature at time 0.