Answer with respect to entropy concepts.
122083
Let's make some assumptions:\n1. The cups are the same size, shape, etc\n2. The amount of hot tea is equal.\n3. Both cups are in the same room at the same temperature conditions\n\nThe laws of thermodynamics state that temperatures will seek equilibrium at a rate proportional to the difference between those temperatures.\n\nSo, by adding milk to one cup, you are slightly lowering the temperature of the tea. So, the milk tea will cool at a slower rate than the other tea because the difference in room temperature and the temperature of the milk tea is less than that of the other tea. \n\nFinal answer: cup of tea
314453
This would not be possilbe.  In the process of digesting and assimilating these essential elements, our bodies create some waste or by-products.  We have to get rid of it somehow.\n\nOur bodies are not Carnot engines - we cannot attain 100% efficiency.  The second law of thermodynamics tells us this (entropy is increasing).\n\nPlus, there are very few pleasures as great as sitting down and passing a nice well-formed turd.
341845
Entropy would increase because when a solid goes to liquid there is more disorder.  They were be more energy because molecules in a liquid move alot faster than in a solid.
128768
Through entropy of heat, the movement of the electrons in the warm surrounding air is transferred to the electrons in the solid/liquid/gas as the electrons accelerate in their orbit around the nucleus the state of the matter changes through the three states I have mentioned.  All of these parameters are leading to the eventual "heat death" of the universe.  Entropy as observed in all systems.
667822
Doesn't sound like you know what ID means let alone irreducible complexity or paley's watchmaker argument to back it up at all and what the hell is your question?  Seems to just be a ignorant claim without any good supporting evidence or logic to back it up.  Keep reading and maybe you'll learn something.
174826
This is an excellent question and is quite tedious for a large number of particles, but there have been some papers written on this question. See http://www.geom.uiuc.edu/docs/forum/hoops_links/energy.html for an interesting aside, http://tracer.lcc.uma.es/problems/thomson/thomson.html for some illustration and http://lanl.arxiv.org/find/grp_q-bio,grp_cs,grp_physics,grp_math,grp_nlin/1/all:+AND+problem+Thomson/0/1/0/all/0/1 for the actual mathematics.
47682
Potential energy is the energy that is available to do work. For example, a ball at the top of a mountain has the POTENTIAL to do energy by rolling down the hill. Entropy is a measure of disorder and randomness. The Universe is in a constant state of randomness. It ties in with potential energy because as the Universe grows more and more random, it spends and utilizes potential energy, reducing the available amount for work.
314297
no, minimum size and maximum energy particles can be derived from uncertanity principle, which is finite. nothing is infinite but only uncertain which cretes an illusion of infinity which is only a distribution of probability, as can be explained by quantum mechanics.
327506
The second law of thermodynamics states that "the entropy of an isolated system not at thermal equilibrium will tend to increase over time, approaching a maximum value."\n\nThe Second Law is a statistical law and thus applicable only to macroscopic systems. When one part of an isolated system interacts with another part, energy tends to distribute equally among the accessible energy states of the system. As a result, the system tends to approach thermal equilibrium, at which point the entropy is at a maximum and the free energy is zero.
117128
The entropy at the exit would be less than the entropy at the nozzle inlet. Adiabatic compression of the steam would generate more friction and overtake natural entropy of the steam. Any adiabatic force would make the entropy less at the nozzle exit.
212665
That's a difficult question to answer, in that we would have to develop experiments to test if our theories were correct. If we had an idealized 'stasis field' (a container or defined zone in which time could be prevented from passing), we might be able to perform experiments to determine whether the things inside were still subject to entropy or whether they were removed from the general entropy gradient. The current prevailing theories hold that the contents of a stasis field (or container) are unaffected by entropy. This leads to the concept that entropy might be stripped from the events which cause it and transferred from object to object or from place to place, leading to unexpected catastrophes for no apparent cause. Once more, these things are difficult to determine.
541345
1) heinsenbergs uncertainity principle\n2)paulings exclution principle\n3)photons(i think so)
197700
Density Matrix.\n\ndiagonal part + nondiagonal part or\nstate probabilities + transition probabilities.
261899
increased entropy is probably your best bet
87967
I think second law of thermodynamics is that physical systems tend to evolve toward states of higher entropy that is toward more disorganization.\n\nIt's not a law in a conventional sense since although rare, and unlikely, something can go from a state of high entropy to one of lower entropy.\n\nIf you toss index cards numbered one through 100 and toss it into air, it's most likely for the cards to land in a random order. But if you toss it enough times, it can land in perfect order, i.e. one to 100.
539294
1 expansion\n2 contraction\n3 boiling point\n4 melting\n5 compression\n6 vaporisation...;
607614
For an adiabatic process dq=0\ntherefore dw=dE\nFor ideal gas E is only a function of temeperature(Eg. 3/2 RT for monoatomic gas)  \nUsing this you can find the end state of the system i,e end temperature , volumes,pressure.\nFor calculating change in entropy you need to understand that entropy is a state function that is it depends only on end state  and not on the path.\ndS=dq/T\nYou can always take any path between the two states integrate and find the change in entropy even if that path is not adiabatic.\nAlso you could use following equation to find final state\nPV^(gamma) =constant \nfor a adiabatic process\nwhere gamma=Cp/Cv