God dammit, now I'm going to need to start doing the calculations for all this, I want to know how much energy is being outputted as heat and how much as usable work. Unless this has been done before? Anyone know if someone's measured the temperature change rate of a CPU and calculated out the waste work?
The more I'm reading this the less it makes sense. Let's take the first part, In-Out=Accumulation. There is none and you mentioned it briefly - this is a steady-state system. I hate arguing by definition but here it is, a steady-state system by definition is a system that doesn't accumulate or lose energy over time. Ie. Edot[SUB]in[/SUB]=Edot[SUB]out [/SUB]or put simply Power[SUB]in[/SUB]=Power[SUB]out[/SUB]. Not to mention that accumulation of free energy in a system is explicitly dis-allowed because of the properties of entropy. Next, the CPU does not move air. The fan does and that's a completely separate system to examine. Yes, I suppose you could include it in the energy balance equation for the CPU insofar as the temperature of the CPU is being managed and therefore the entropy of the system is radically different between the two cases: CPU w/ fan and CPU w/o fan. However, it's only relevant when taking into account the power lost through inefficiency. The CPU fan does not effect the work done by the CPU, remember the Edot[SUB]out[/SUB] is the same and the Edot[SUB]in[/SUB] and the only difference between case 1 and case 2 is the temperature of the CPU. Lastly, the work done on the surroundings is heat but we need to define the surroundings first. For simplicity, I'm going to say the surroundings are the fan-side contact with the CPU, the radiator on the fan, and then the surrounding air. You're right, the kinetic energy is negligible but I'm going to stick my neck out and also say the thermal energy is negligible as well (done on the environment [remember this is a different system than the CPU {the energy is not at all insignificant in the CPU system}]). The specific heat of metals and of air are pathetically low compared to, say, water. Air is about 1kJ/kgK and aluminum is about .91. Per kilogram. CPUs are weighed in grams.
Over time or not, it doesn't matter. Time will multiply out you can use rates or total values for the general picture here. I think it should be pretty easy to see why the whole computer system is both the easy system and the important system to worry about if you want to know the impacts on say, the room the computer is in. Anyway, the point was to show that accumulation is zero and that there's no other significant form of work being done on surroundings showing why basically all of the box's consumed power has to come out as heat. Right? What might accumulate? Nothing really, so you're left with energy in - energy out = 0. In = power coming out of the wall. Out = heat and kinetic energy of the air it pumps through. What else is there? If you want to make the balance over the CPU your answer has to be much the same. Now you do have some power that will flow through the CPU. But any power consumed by operation of the CPU has to come out as heat, with the mechanical switch analogy illustrating that. When it comes down to it a CPU is basically a little electric heater that can manipulate some data for you (and "data" or "answers to calculations" have no intrinsic energy value...we can go into that if you want because I have thought of some neat analogies!). You last paragraph is getting a little "what?", maybe you're not stating it quite how it is in your head. Power figures typically quoted by companies are for TDP, thermal design power. So that 220w is what AMD has said a cooling system needs to be able to remove from the CPU. That's the amount of heat which has to be removed or else the CPU absolutely will accumulate heat until it melts its packaging and fails. All of that has to end up going to the air. Neglecting the heat capacity of the air means you're neglecting all of your cooling capacity which is very silly. Unless you're like running tap water though your system to a drain or something really odd. Especially of note is that you're typically pushing at least 10CFM through your case and rather significantly increasing its temperature. The air adds up.
You're confusing the heat generated with the work done. They are not at all the same thing. Let's stick to mechanics to explain: if Earth had no atmosphere andyou dropped something from orbit, it would not accumulate heat. The work done would be all kinetic with no resistive forces (no drag). However, if there was an atmosphere there would be the work done AND heat generated due to friction therefore it's not the work itself that generates heat, it's the resistance to the gravitational force that generates heat. Extending the analogy to the CPU, any energy consumed (used) by the CPU is not heat, it's work. The associated temperature change is solely caused by heat and that's due to resistive forces not the CPU itself calculating something out. Also, the kinetic energy of moving air is an entirely separate system. The fan moves air. The CPU does not. In this case, especially because it's micro-circuitry, that's a significant difference. Of course they have an energetic value. The CPU wouldn't consume power (produce output work) unless the thing it calculated required energy to calculate. You're right about the last paragraph, I didn't explain correctly. The energy due to heat is probably no where near 100 watts. It's most likely much smaller. For the CPU, the heat generated is a significant factor when calculating the energy balance because the ratio of heat:work is significant. In the fan, it's not that important because the fan does significantly more work to remove heat than there is actual heat (entropic principal - you need to put in work to remove heat from a system).
No, I'm not. You're leaving out what happens at the end. Regardless of frictional or nonfictional atmospheres the meteor hits the ground deforms material and generates heat. The work of moving the electrons around the CPU generates heat, there isn't any deformation or other expenditure of energy being done. They are not moving up in potential energy and staying there and they are not given kinetic energy that stays permanently. The CPU does not undergo significant physical changes in structure. The CPU at the end is the same as it was in the beginning. No, there's a critical difference between information itself and the methods of storing and manipulating it. You can move stones around on a surface to store data. They can be big or small. The energy required to move a stone is vastly different between large and small stones (at the same speed). Smaller stones, faster calculations or lower energy usage up until you reach the minimum stone size. Think about a flat, frictionless surface: Any arrangement of the stones contains different information but all arrangements have the same energy content. And all movement costs the same energy (no friction means movement of any distance costs the same energy). Calculations are performed though moving them but all of that energy is simply added to move the stone and removed to make the stone stationary in its new position. It's not the data itself but the methods which cost energy. In the case of a CPU we're conducting electrons around in different patterns the work done to flip the switches becomes heat, all of the movement of electrons through the wires generates heat. A little bit of power comes out the end to report what's gone on. It is vastly, orders of magnitude lower than flipping those switches and conducting through those wires. Which is why none of this internal stuff matters at all to the energy balance on your computer box or even the CPU alone. As a black box it sucks up energy but hardly does any appreciable work on the outside world and it doesn't accumulate any potential energy or do any lasting changes to itself which consume energy. So all of the work done is internal yet it leads to no accumulation of energy. It has to end up as heat. The box says 84 or 220w or whatever TDP because that's how much heat the cooling system needs to be designed to handle under some manufacturer-specified usage conditions. So while actual consumption may be higher in a hypothetical sense...that's not what that number is talking about.
At what point does the work done by gravity generate heat when something falls in a frictionless environment? The Earth is a dissipation tool, like the fan it doesn't need to be included in the system to fully describe the energies within in. The work done by gravity has nothing to do with the deformation of the Earth. Between one moment and the next, the systems are different. The deformation of the Earth is due to the kinetic energy of the ball after all the potential energy of the object has changed. Two things disprove this statement. E=mc2 and F=ma. Larger and larger stones have a larger energy content because they have larger mass. Information is energy. The more information you handle, regardless of medium, the more energy you're going to need to express it. The larger stone by its nature cannot equal a smaller stone. Larger stones even on a frictionless plane require more energy to move. To move a small stone it requires less energy. You're contradicting yourself when you say they take the same energy to express some information. It does not. Also, on a frictionless surface their is no heat generation, again pointing out that work does not mean heat. In it's simplest form, a steady-state energy balance takes the form of W[SUB]in[/SUB]+Q[SUB]in[/SUB]=W[SUB]out[/SUB]+Q[SUB]out[/SUB] where W=work and Q=heat. These things are mutually exclusive. Heat is not work. Heat is waste. Especially in this case where there is no Q[SUB]in[/SUB]. You're contradicting yourself. You can't say the internal stuff matters and then turn around and say that the external stuff doesn't matter either. You're overlapping systems together. The CPU does perform work on the environment. The environment does not perform work on the CPU. I don't understand what you mean when you say that there is no accumulation of energy due to work. What has to end up as heat? The work going in? How, if there was no work coming out where would the calculations occur? Heat is chaotic, it is waste, it is not structured calculation.
All the work done on a freefalling object which is gained as kinetic energy, if the object is your system, has to go someplace when the object stops. Consider the "never stopping" case has nothing to do with our system. If we're talking about the system of both the earth and the falling object, the net energy change is obviously zero when something falls. But it's changing from potential energy, when the object is far away, to kinetic as it falls, to some kind of deformation of the earth and object and heat when it collides with the surface. Pick one or the other, object or earth, or the whole combined system. The falling and subsequent collision, in a real system (in vacuum or air) results in some energy going to deformation of one or the other, earth/ball, and a higher average temperature of both objects. If the ball is our system the ball falls and gains energy through gravitational work and then collides and loses that kinetic energy to work performed on deformation of the earth, the object and raising the average temperature of both things...or else it's just going to bounce up and down ad infinitum. The bouncing end obviously doesn't apply to real systems. This is irrelevant. We're not adding or subtracting mass here and the physical nature of the stones is completely unimportant outside of mass. The placement of the stones is the data we're interested in and the energy required for the placement of the stones it the energy for the calculation. You have to both speed them up to move and slow them down to stop them in the new configuration. Neither of which can be perfectly efficient. Except I didn't say they take the same energy. I pretty explicitly say the energy is different. Moving large things at the same rate as small things takes more energy. There is a minimum for the amount of heat which must be generated from moving the "smallest stone" and it's explained by the Landauer Principle but I'm not sufficiently prepared to try and explain it. The whole point of that equation is to show that energy is conserved, being that's the unit of both work and heat. There's no Qin, true. There's also only the tiniest Wout. So Win=Qout. How else would an electric stove work? Also, to say heat is simply waste is sketchy at best...if there's a temperature difference to be had heat flow can be made into "useful" work... I'm trying to explain the internal stuff because i thought it might help illuminate the large picture, given you stared talking about the internal aspects. We can easily talk about a computer with no moving parts at all, no active cooling and even no external output but it still does calculations and stores the results internally on an SSD or something. If we put it in a calorimeter (and had good data on it's composition) we'd see that all the electrical power that went into it ended up raising its temperature. To draw the energy balance around a computer and see why basically all the electricity that goes in has to come out as heat shouldn't require an in depth explanation of how a transistor or anything inside of the machine works. What work? What work does it do that comes even close to its power draw? Is it lifting objects? Compressing gasses? Powering external motors? Driving chemical reactions? It just has a little bitty output signal. It does nothing on scale with power consumption. You have to plug them in...that's not not providing energy. You plug it into a wall. There is energy going into it. It reaches a steady state temperature, it does not change chemically or physically in a permanent matter. It does not do significant work on anything outside of itself. The only way for the energy balance to close is for it to radiate heat.
I think we need to start over here. We're confusing systems by this point. As a basic starting point for the gravitational example: I'm breaking it into two systems the first being the ball from rest until the moment before impact and the second the moment of impact until deformation and beginning of rebound (assuming partially elastic collision). Can we agree to use that dichotomy to define the total system? I choose this method because you can "see" all the energy is accounted for. (Potential to kinetic in system 1, kinetic to potential and (deformation, sound, heat, etc). In the CPU example I'm considering three systems: the first is the electrical power delivered to the CPU and the computation, the second is the electrical power delivered to the fan to displace air, the last is the heat transfer between the contact point (fan-side), radiator, and surrounding air in the case. Is that acceptable?
On a budget AMD is the way to go, not to mention all your gaining really from Intel is a cooler temp. If your looking for great gaming on a budget the 8-core 8350 which is what i'm running and i never exceed 40% usage of my cpu and I put it through hell. I bought mine from Microcenter for a little over $200. Intel can't match that price because your paying mostly for the logo.