a) The change in the internal energy of a system that absorbs 2,500 J of heat and that does 7,655 J of work on the surroundings is _________ J.

b) The ΔE of a system that absorbs 12.4J of heat and does 4.2 J of work on the surroundings is __________ J.

Respuesta :

Answers:

a) 10155 J

b) 16.6 J

Explanation:

According to the First Law of Thermodynamics we have:

[tex]\Delta E=Q + W[/tex]

Where:

[tex]\Delta E[/tex] is the total change in the internal energy of a system

[tex]Q[/tex] is the heat exchanged between the system and its surroundings

[tex]W[/tex] is the work done on the system

a) In this case [tex]Q=2500 J[/tex] and [tex]W=7655 J[/tex], hence:

[tex]\Delta E=2500 J + 7655 J=10155 J[/tex]

b) In this case [tex]Q=12.4 J[/tex] and [tex]W=4.2 J[/tex], hence:

[tex]\Delta E=12.4 J + 4.2 J=16.6 J[/tex]

ACCESS MORE