Quick n Easy Chemistry Question?? for TEN POINTS!!!?

[James M.]

A system releases 125 kJ of heat, and 104 kJ of work is done on it. Calculate the (delta symbol) E. (delta symbol) E is the change in internal energy. I got 229. But I wasn't sure. The other options are -229, -21, 21, or -125.

[Paul]

I'm pretty sure it's -21, because when a system releases heat, that means its exothermic and is negative, and the work being done on the system, so that's an endothermic process, thus -125+104=-21

[yankee_fan907]

Delta E = q + w, where q is the heat of the system and w is the work of the system.

In this particular problem, q is equal to 125 and w is equal to 104. But because we always look at a system's overall energy change and not the surroundings', we know that the 125 must be negative. This makes sense because the system is releasing the energy and no longer has it. The work done on the system is positive because the energy is being transferred into it. If work was done by the system, then it would be neative.

So -125 + 104 = -21 = Delta E

I hope this helps, and explains it too!

[yankee_fan907]

Delta E = q + w, where q is the heat of the system and w is the work of the system.

In this particular problem, q is equal to 125 and w is equal to 104. But because we always look at a system's overall energy change and not the surroundings', we know that the 125 must be negative. This makes sense because the system is releasing the energy and no longer has it. The work done on the system is positive because the energy is being transferred into it. If work was done by the system, then it would be neative.

So -125 + 104 = -21 = Delta E

I hope this helps, and explains it too!