Question

108J of work was done on a closed system,during this phase of the experiment,79J of heat energy was added to the system. what was the total change in the internal energy of the system.

Answer 1

The internal energy of system is 29 joule .

Step-by-step explanation:

Given as :

The amount of work done = Δ W = 108 joule

The heat energy added to the system is Δ Q = 79 joule

Let The change of internal energy = Δ U

Now ,

The change in internal energy = work done by system - heat energy added to system

I.e Δ U = Δ W - Δ Q

Or, Δ U = 108 joule - 79 joule

∴ Δ U = 29 joule

Hence The internal energy of system is 29 joule . Answer

Answer 2

Internal Energy increases by 187 J

Step-by-step explanation:

Consider the system on which experiment is done.
`108\text{ }J` of work was done on the closed system. All this work will contribute to rise in Internal Energy.

`79\text{ }J` of heat energy was added to the system. This heat is also a form of energy added to the system. Hence, it adds to the Internal energy.

`Q=\text{ Heat added to the system }=+79\text{ }J`

`W=\text{ Work done on the system }=+108\text{ }J`

`\Delta U=Q+W=108+79=+187\text{ }J`

∴ Internal Energy increases by 187 J