I heard a claim that due to holographic principle, the surface area of the cosmic horizon corresponds to the universe's total entropy.
As such the initial state had zero surface area and later expanded.
Given this, I wonder whether any increase in entropy (such as producing heat by means of electric power) causes the universe to expand?
The wikipedia article about the holographic principle does not suggest, that this is well-accepted and founded theory :)
The ideas seem to be somehow based on the concept of the Bekenstein-Hawking entropy for a black hole. This is supposed to be the entropy of a black hole from outside (according to the no-hair-theorem the black hole has only three numbers that describe it, so any information about the things that fall into it is lost, and simultaneously it is supposed to have a high entropy proportional to the area of the event horizon - modern physicists are crazy!).
In this case, to get back to your question, the event horizon is not influenced by anything happening inside the black hole (obviously, this would imply information getting outside).
The argument against the radius of the universe depending on the events inside it is analogous: it was expanding with more than $c$ and is too far away to be affected by this.
Well if I go in a simple way, it's all the interaction of matter with each other on both the quantum and large scale and the energy released in this process continually increases the entropy of the universe and thus expanding the cosmic horizon. But the major quantum fluctuations are very very very small i.e. something around $10^{-123}\,.$ Now scientists are wondering how these minute fluctuations result in the increase in the entropy.
This is a very good question. But before I attempt to give you some of the details, the increase in the entropy is not what causes the universe to expand but is rather a consequence of the expanding universe. In fact, to understand why the entropy of the universe was so low before the inflationary epoch is an open question. Based on this question, I am assuming you are interested in de Sitter spacetimes.
Unlike flat space and anti-de Sitter spacetimes, de Sitter is a bit more nontrivial to deal with. In trying to compute the entropy and other thermodynamic quantities like for black holes, we can proceed by computing the quantity $$\frac{dS}{dE} = \frac{1}{T}$$ which for de Sitter is $$\frac{dS_{dS}}{dE_{dS}} = \frac{1}{T_{dS}}.$$ Let us stick to the case of 2+1 dimensions for now. In the case of 2 + 1 dimensional de Sitter spacetime, we have no good understanding of the meaning of $E_{dS}$ because once we fix the cosmological constant value, unlike the case of Schwarzschild spacetimes, we do not get a family of metrics as the solutions but rather only one solution which indicates the presence of a cosmological singularity.
So to get around this, we start with a Schwarzschild like object in $dS_3$
$$ds^2 = -(1- 8GE - r^2) dt^2 + (1- 8GE - r^2) ^{-1} dr^2 + r^2 d\phi^2 $$ where $1 - 8 G$ is the square of the radius of the de Sitter horizon or $r_H^2$. You can analytically continue this to the complex plane and compute the temperature via the periodicity of the Green's functions solution and you find that the entropy is $$T = \frac{\sqrt{1- 8GE}}{2\pi}.$$ Plug this back into the differential equation for the entropy and using the value of $r_H$, you find that the area is given by $S = \frac{A}{4G}$.
This is also achieved via a different method i.e. via the algebra of global diffeomorphisms in $dS_3$ similar to the Brown-Henneaux approach in which case you compute the central charge of the field theory that lives on the boundary and compute the entropy via the Cardy formula. See the paper by Balasumbramanian et. al.
For a conjecture on higher dimensional de Sitter spacetimes, see this paper by Bousso. For proof of this bound in the classical and quantum limit see this paper and this paper too!
The holographic principle is a very broad and generic conjecture and its proof will take a lot more scientific effort. But maybe we're on the right track.
