Having checked this, you’re totally correct.
Here’s a short Python program that checks this empirically.
import numpy as np
n = 100
b = np.zeros(n)
r = np.ones(n)for ib in range(n):
for ir in range(n):
old_b = b[ib]
old_r = r[ir]
new_b = new_r = np.mean([old_b,old_r])
r[ir] = new_r
b[ib] = new_bprint(np.mean(b))
A graph of the scaling of the max blue temperature with respect to number of divisions is here:
https://imgur.com/a/r4hdaJD
When T is 1000 you can get the blue to 0.98 of what red was initially… so I imagine in the infinite limit you can get all the way there.
This seemed weird to me, so I looked at the final temperature distribution of all of the blue blobs, as increases. I’ve plotted them here (blue lines have being small; red lines have being big). You can see that as the number of blobs becomes very big, the fraction which have temperature less than 1 falls… (i.e. knee of graph moves to the right...)
Treat the blue clay (mass 1, temperature T) as a single lump. Feed it infinitesimal lumps of red clay (mass dX, temperature TR). After each infinitesimal feeding, the temperature of the blue clay changes by
dT=T+TRdX1+dX−T=−TdX+TRdX
(Final equality comes from series expansion).
Then you can integrate: ∫Tfinal0dTTR−T=∫10dX.
i.e. log(TRTR−Tfinal)=1
The final solution is:
Tfinal=TR⋅(1−1/e).
It’s worth saying that this is the most efficient way to transfer energy from red to blue because each feeding step is thermodynamically reversible.