= Solution
For <speed> distribution $f(v)$ normalized to one and isotropic directions, the incident number flux is
$$
J=n\int_0^\infty vf(v)\,dv\,
\frac1{4\pi}\int_0^{2\pi}d\phi\int_0^{\pi/2}\cos\theta\sin\theta\,d\theta
=\boxed{\frac14n\overline v.}
$$
The positive normal <velocity> weights the population reaching the wall. Consequently its mean translational <energy> is
$$
\langle E\rangle_{\rm hit}=\frac m2\frac{\int_0^\infty v^5e^{-av^2}\,dv}{\int_0^\infty v^3e^{-av^2}\,dv}
=\frac m2\frac{a^{-3}}{(2a^2)^{-1}}
=\boxed{2k_BT},\qquad a=\frac{m}{2k_BT}.
$$
The bulk value $3k_BT/2$ would omit the arrival weighting.
For the two vessels, let $r=\sqrt{T_1/T_0}$, with subscripts h and c for hot and cold. Equal opposing <effusion> rates give $p_h/\sqrt{T_1}=p_c/\sqrt{T_0}$. Number conservation in equal, fixed volumes adds $p_h/T_1+p_c/T_0=2p_0/T_0$. Hence
$$
\boxed{p_c=\frac{2p_0r}{1+r},\quad p_h=\frac{2p_0r^2}{1+r},\quad
p_h-p_c=\frac{2p_0r(r-1)}{1+r}.}
$$
Both final <pressures> must be determined; holding the cold <temperature> fixed does not hold its <particle number> fixed.
For aperture <area> $A_h=\pi d^2/4$, the common one-way flux is $J=p_c/\sqrt{2\pi m k_BT_0}$. The translational <heat> transfer from hot to cold is therefore
$$
\boxed{\dot Q=2k_B(T_1-T_0)A_hJ.}
$$
This is the total <heat> transfer for a monatomic gas with no active internal modes, neglecting radiation and conduction through vessel walls. For a molecular gas, add $A_hJ[u_{\rm int}(T_1)-u_{\rm int}(T_0)]$ for its <internal energy> per molecule. The maintained unequal <temperatures> describe a stationary nonequilibrium state even though the net particle current is zero.
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