Which one true in First law of thermodynamics: $Q = \Delta U \pm W = \Delta U \pm p\Delta V$ or $\Delta U= \Delta Q + \Delta W $? Which one true in First law of thermodynamics: 


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*$Q = \Delta U \pm W = \Delta U \pm p\Delta V$? (where $\Delta U$ is change of internal energy, $W$ work made by system and $Q=cm\Delta T $ heat made by system.)   

*or $\Delta U= \Delta Q + \Delta W $? (where $\Delta U$ is change of internal energy, $W$ work made by system and $Q=cm\Delta T $ heat made by system.)
First one from one the finnish physics texbook another is from here.
 A: It is pretty much a matter of convention regarding who is doing work on whom. For me the most conceptually clear picture is the wikipedia version,
$$\Delta U=\Delta Q+\Delta W,$$
i.e., that the change in the internal energy of the system equals the heat delivered to it plus the work performed on the system. (Note, however, the difference from what you quote!) If one takes $\Delta W$ to be the work performed by the system then its sign will change, but this does not of course change the physical content of the law. If in doubt, put it in words! Once you're clear on what each symbol means the signs will follow automatically.
A couple of caveats, though: note that $Q$ as such is a misleading term. One can only assign heat quantities to processes, which is emphasized by the notation $\Delta Q$.
A: What's true is that it's law of conservation of energy. 
It particularly states: Q= Del(U) + W, 
U: internal energy. 
That is, the total energy given to a system does two things, first it makes the system do the desired useful work and second it changes the internal energy of the system. 
The work can be positive or negative according to W=q(dv). 
Example: If the gas in piston expands and when the gas in the piston is under compression then the work done would be positive and negative respectively. :) 
A: For me, it's
delta(q) = delta(u) + w 

delta(q) is the change of heat of the system, delta(u) is the change of internal energy of the system and w is the work done by the system. delta(w) just doesn't make sense for me since w by itself means the change in energy. Also, writing it this way allows you to use parts of the statement in other places as well. For example, change in entropy delta(s) = delta(q) / T. And total work done by the system in an adiabatic process w = nRTln(v2/v1)
A: The previous answer, in my opinion is half-baked. The use of the delta term is quite misleading, as pointed out in the comments. Now, according to almost every leading textbook and my Professor, who is also a leading author, the I Law of Thermodynamics can be stated as:

Q = ΔE + W

where Q is the heat transfer across the system boundary, W is the work transfer across the system boundary and ΔE is the change in energy of the system. Here, the energy of the system 'E' includes the Macroscopic Energy mode(Kinetic and Potential Energies of the bulk) and the Microscopic Energy Mode(Internal energy that included the translational kinetic energy, rotational kinetic energy, vibrational potential energy, chemical energy etc. of each molecule).
The sign of Q and W depend on the sign convention used. Hence, you don't need to include the +- signs in the equation as it will only tend to confuse you!
The above statement gives a whole picture of the I Law of Thermodynamics for a 'Closed System' undergoing a 'Change of State'.
