How can I explain what a kilogram is using Planck's constant? I want to understand what $1\ \mathrm{kg}$ represents. For example: I know that $1$ second is equal to $9\,192\,631\,770$ transitions from the microwave radiation that a cesium-133 atom (at $0\ \mathrm K$) emits, if it's excited just right. I can imagine that. I can see how you would "count" these 9 billion transition until you know, that exactly $1$ second has passed.
Now I would like to know if there is a similar explanation for the kilogram. 
I understand how Planck's constant has be redefined using methods such as the Kibble balance. I would like to know how I can explain what $1\ \mathrm{kg}$ is using $h$. Here is what I've got so far:
Knowing that $E=hf$ and $E=mc^2$, if both of those energies are equal, this gives $m=\frac{hf}{c^2}$. So if we want to know what $1\ \mathrm{kg}$ is, we find the frequency $f$, that gives $\frac{hf}{c^2}=1\ \mathrm{kg}$, which would be $1.3564 \times 10^{50}\ \mathrm{Hz}$. 
What does this frequency represent? Is it the frequency of light that you would need to "push" an object with a force equivalent to the weight of $1\ \mathrm{kg}$?
Sorry if my thinking is completely off.
Edit: the answer of the question What are the proposed realizations in the New SI for the kilogram, ampere, kelvin and mole? explains in detail how the new units get defined and what their relations are, but does not give a satisfying explanation as to what e.g. a kilogram represents.
 A: Quoting this excellent thought experiment from the article An atomic physics perspective on the new kilogram defined by Planck’s constant by Wolfgang Ketterle (thank you wcc!):

The  new  kilogram  may  be  understood  as  the mass difference of $1.4755214 \times 10^{40}$ Cs atoms in the ground state versus the same number in the excited hyperfine state or as the mass of $1.4755214 \times 10^{40}$ photons at the Cs hyperfine frequency trapped in a microwave cavity. The numerical value 
  $$
1.4755214 \times 10^{40}\,\text{kg}^{-1} = \frac{c^2}{h \cdot \nu_{Cs}} = \frac{299\ 792\ 458\,\text{m}^2\text{/s}^2}{6.626\ 070\ 15\times 10^{-34}\,\text{kg m}^2\text{/s}\ \cdot 9\ 192\ 631\ 770\,\text{s}^{-1}}
$$ is fixed through the definitions of $h$, $c$ and $\nu_{Cs}$  and has no uncertainty.
In a thought experiment, one could measure out 1 kg of any substance by having a mechanical balance where the substance and the ground state Cs atoms on one side are compared with the Cs atoms in the excited hyperfine state on the other side of the balance.

So this is it! I think this is probably the best way to think about what this new definition of the kilogram actually represents: It's the difference in mass of a bunch of cesium atoms in one energy state versus another.
A: Energy and mass are equivalent. The two equations for energy you start with are equating the energy contained within 1kg of matter and the energy of a photon having frequency f. So, a single photon with with a frequency of 1.3564e50 Hz has the energy equivalent of 1kg of mass. As you might imagine, that is an extremely energetic photon, many orders of magnitude past where we've stopped naming parts of the electromagnetic spectrum - it'd be a super-duper-ultra-high-energy gamma wave.
A: The definition is pretty confusing but boils down to: $$ h = { 4 \over pn^2 }* {gv \over f^2} * m $$
where h =Plancks constant
p is an integer
n is the number of Josephson junctions in the Kibble balance and is a large number of the order of 10^5.
g = local acceleration due to gravity  
v = velocity of the cylinder in Kibble Balance when running in velocity mode
f = frequency of microwaves in Josephson junctions
m = mass
and h is defined as exactly 6.62607015×10−34 kg⋅m2⋅s−1 
There is a small movie explaining this in detail. This is the best general description I could find.
