Schoenfeld: Still fails to bull**** his way to a nobel prize,

We fill a box with a radioactive material which randomly emits
radiation. The box has a shutter, which is opened and immediately
thereafter shut by a clock at a precise time, thereby allowing some
radiation to escape.

Let's see if I understand this. You plan to use radioactive decays,
which have measured decay rates in agreement with quantum mechanical
calculations to try and refute quantu mechanics? Doesn't that seem a bit
illogical? Wouldn't it be simpler just to find a radioactive decay
that violates the uncertainty principle and save the work involved
in creating a rube goldberg experiment that relies on delusional
handwaving rather than hard numbers from an experiment that can
be performed, in principle?

So the time is already known with precision.

It's always useful to begin a string of handwaving platitudes
with a statement that assumes the result by virtue of not limiting
yourself to a real physical process.

We still want to measure the conjugate variable energy precisely.
We measure the energy by weighing the box before and after.

The uncertainty in the energy due to only classical thermodynamic
fluctuations (no quantum mechanics involved) is,

\Delta E = sqrt(kC_v) T, C_v = heat capcity at constant volume

The mass if the box before and after a shutter click is given by
m_f = m_i - \Delta m. The change in mass has to exceed the classical
fluctuations in order to determine the mass has changed, so

\Delta m sqrt(kC_v) T/c^2

The termperature of the box must be greater than the surrounding
environment, or else radiation will enter the box with equal probability
that it leaves the box. So take T to be some T 300 K, like 1000 K.
For C_v, an order of magnitude estimate is sufficient and for a
monatomic ideal gas, C_v = (3/2)nR. If n is 1 mol, then C_v = R/a_avag = k,
where a_avag is the avagadro constant.

sqrt(kC_v) T/c^2 = sqrt(3k^2 a_avag/2) T/c^2

The smallest energy you you can measure in any shutter click that could
differentiate between a change in mass and a fluctuation from just
classical considerations in 1 mol of gas is,

\Delta E of about 81 GeV or a mass of about 1.5 x 10^-25 kg

To violate the uncertainty principle, you therefore need shutter
speeds that violate the uncertainty principle which is easy to
determine:

\Delta t hbar/(\Delta E)

Meaning you have to make measurements of \Delta E = 81 GeV for intervals
of \Delta t 10^-26 sec.

So, what does that mean in terms of the power being dissipated by
the box? Well, P = dE/dt, so P = 1.3 x 10^19 watts for 1 mol of
nuclei in your box. That's a rather substantial rate for electromagnetic
radiation and since that won't happen for a mol of gas at 1000 K, you
are out of luck for E&M radiation.

But you can always try alpha decay so that the 81 GeV is in the form
of low energy alpha particles. 20 each 10^-26 seconds to be precise for
1 mol of pure radon gas. You can also drop the temperature back down,
but that doesn't help a lot. How many radon decays occur each 10^-26
seconds, on average for 1 mol of pure radon gas? The half-life of radon
is 3.8 days. When you're through doing it the hard way, see if you can
figure out why this should have been obvious from the start.

Unfortunately, regardless of how you do it, you'll need a detector
that violates the uncertainty principle to count the alpha particles.
Violating the uncertainty principle with apparatus borrowed from
a universe with different physical laws doesn't count, unless you
can demonstrate that you borrowed such equipment and it works in
this universe.