## First Law Thermo

$\Delta U=q+w$

Emilie Hoffman 1E
Posts: 55
Joined: Fri Sep 29, 2017 7:04 am

### First Law Thermo

Is q(system) + q(surroundings) = 0 another iteration of the first law, because that's what I have written in my notes. And if so, can someone explain how?

Tia Tomescu 2D
Posts: 42
Joined: Fri Sep 29, 2017 7:06 am

### Re: First Law Thermo

No because the first law is deltaU = q + w and but equation you have is only related to q.

Jessica Jones 2B
Posts: 80
Joined: Fri Sep 29, 2017 7:04 am

### Re: First Law Thermo

the first law of thermo is that the internal energy of an isolated system remains constant. From this concept, we get deltaH= q + w.
The reason why qsys + qsurr= 0 is because when energy leaves a system, the surroundings gain energy(as energy is never created nor destroyed) and so from this can use them in relation ship to each other and solve for things. qsys = - qsurr

Kyung_Jin_Kim_1H
Posts: 53
Joined: Thu Jul 27, 2017 3:00 am

### Re: First Law Thermo

q(surr) = -q(sys) is correct, but it is not another accurate explanation of the rate law because we'd be ignoring the work component in internal energy.

Troy Tavangar 1I
Posts: 50
Joined: Fri Sep 29, 2017 7:04 am

### Re: First Law Thermo

No, the first law involves work and is deltaU= q +w