Qsystem+Qsurr=0?

[Max Hayama 4K]

Can anyone explain the reason behind the negative or positive sign when calculation delta H? I understand it's negative because it exothermic and vise versa, but I was confused when Professor Lavelle discussed the qsystem and q surrounding.

[marisaimbroane1J]

The equation is based off of the fact that if the system gains heat, it has to take it from the surroundings-- resulting in 0 (k)J of heat "lost". The same is true if the system gives off heat (-); the heat goes to the system (+), again the heats "cancel" out, resulting in the zero seen in the equation.

[Nico Edgar 4L]

The equation is based on the fact that the surrondings can either lose or gain energy due to the reaction happening in the system. So if the reaction in exothermic, it releases energy and heats the surrondings, so the Q of the system would be negative and the Q surr would be positive. We assume our systems to be perfect so the amounts would be equal.

[Yiting_Gong_4L]

The Qsystem and Qsurrounding cancel out for the system to be perfect. Therefore the equation would equal 0.

[Carine Tamamian 2B]

If one loses heat, that heat will be gained by the other. The loss will equal the gain, which makes sense.

[Sophia Ding 1B]

It's kind of like one of the laws of thermodynamics which I know we haven't covered yet but might help explain the concept of it; energy can never be created nor destroyed. With that in mind, the heat of the system + the heat of the surrounding would equal 0, because as one loses heat such that the other gains the heat, they "cancel" out since this heat energy can't be destroyed in the universe but rather is just transferred to one another.

[904914909]

The system gains the heat of the surroundings in some reactions but the system loses the heat to the surroundings in other reactions so the net change in heat is 0 which is why Qsyst + Qsurr = 0.