If a system generates 1300KJ energy to its surrounding what would be the change in temperature?

First, It's always necessary to report the heat power or the energy and the heat dissipation time.

The reason is that in normal situation, you will always have heat transfer ( gain or loss) to the surrounding of the container.

Nevertheless, and assuming an ideal situation where the heat transfer is null, you must define the following:

- What is the material and mass of the surrounding you are talking about, i.e. if it's a well tight room, then the surrounding will be air, the mass will depend on the Air (Room) volume and the Air specific weight (depending on the room ambient temperature).

- You need also the "specific heat" of the surrounding. The specific heat of Air depends on the temperature, the humidity and the pressure.

If the surrounding is water or any other material, then

Delta T = E/m.c

E: Energy is Calorie, knowing that1 cal =4.182 Joule

m: weight in gram

c: Specific heat of the materials

There are too many factors and conditions that may affect the result