Question

If you do 1,000 J of work on a system and remove 500 J of heat from it, what is the change in its internal energy?

-1500 J, -500 J, 500 J, or 1500 J

thank you!!

Answer 1

here we have to apply the first principle of thermodynamics:
\[\Large Q = L + \Delta U\]
where \Delta U is the internal energy change

Answer 2

okay!! so 1000+500? so the solution is 1500 J ?

Answer 3

more explanation:
since do the work L on the thermodynamic system, then L=-1000 J, namely it is negative, so we have:
500=-1000+ \Delta U
then:
\Delta U = 1500 Joules
so your answer is correct!

Answer 4

oops.. since we do the work on the thermodynamic system...

Answer 5

oooh yay!! :D

thank you:)

Answer 6

thank you!! :)