Waxwing

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_b
print(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... (read more)

Replying toSleep math: red clay blue clay

Waxwing5y

Sleep math: red clay blue clay

Treat the blue clay (mass 1, temperature T) as a single lump. Feed it infinitesimal lumps of red clay (mass , temperature $T_{R}$ ). After each infinitesimal feeding, the temperature of the blue clay changes by

$d T = \frac{T + T_{R} d X}{1 + d X} - T = - T d X + T_{R} d X$

(Final equality comes from series expansion).

Then you can integrate: $\int_{0}^{T_{f i n a l}} \frac{d T}{T_{R} - T} = \int_{0}^{1} d X .$

i.e. $log (\frac{T_{R}}{T_{R} - T_{f i n a l}}) = 1$

The final solution is:

$T_{f i n a l} = T_{R} \cdot (1 - 1 / e)$ .

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.

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