# Decrease in internal energy

1. Feb 17, 2015

### brake4country

1. The problem statement, all variables and given/known data
In which of the following cases does the internal energy of the described system increase:
(I) A clay ball is dropped and sticks to the ground
(II) Hydrogen and oxygen react exothermically to form water
(III) A car is driven at constant speed (consider the effect of air resistance)

2. Relevant equations
NA

3. The attempt at a solution
When I look at I and III, I think of kinetic energy and potential energy (if elevated). But then when I think of internal energy, I think chemical reactions. How can a macroscopic object such as I and III have an increase in internal energy?

2. Feb 17, 2015

### Staff: Mentor

If a chemical reaction is carried out exothermally in a constant volume reactor, does that mean that heat is added to the reactor or that heat is removed form the reactor to return the temperature of products to the original temperature of the reactants? So what is the sign of Q? What does that tell you about the sign of ΔU?

If a falling ball of clay has kinetic energy when it hits the ground and the deformation of the clay takes place without any heat transfer from the clay to its surroundings, does the ground do work to deform the clay? What does that tell you about the sign of ΔU?

Chet

3. Feb 17, 2015

### brake4country

If the volume remains constant when an exothermic reaction takes place, then q would be positive, which means that ΔU would be positive, right? This would mean that heat is added to the system, thus giving the system potential to do work.

KE = W with macroscopic systems, so a ball of clay with kinetic energy has the potential to do work without heat. Please let me know if my logic here is correct. Thanks!

4. Feb 18, 2015

### Staff: Mentor

Exothermic means that q is negative. Heat is removed from the system. How much work is actually done by a material in a constant volume reactor? Is ΔU positive or negative?
The work that the system does on the surroundings is zero, because the displacement of the ground is virtually zero. So the work done by the ground on the clay ball is also zero. Don't forget that, when there is a significant change in kinetic energy, the first law must be modified to $ΔU+ΔKE=Q-W$. So the kinetic energy gets converted into internal energy. So, is ΔU positive or negative (i.e., does the ball of clay get hotter or colder when it goes splat?)

Chet

Last edited: Feb 18, 2015
5. Feb 18, 2015

### brake4country

Okay, a lot of great questions! So, for the first one, no work is done if the volume remains constant. This is an isovolumetric system. If heat were to go into the system, then it would be endothermic and the q would be positive.

For the second question, conceptually thinking about this I would say that the clay would get hotter due to the kinetic energy that it has upon hitting the ground. This would be similar to an exothermic process where heat is removed from the clay in to the surroundings. Is this thinking correct?

Therefore, q is negative then ΔU must be negative also. Correct?

6. Feb 18, 2015

### Staff: Mentor

Right. Since it's exothermic, q is negative.

No. q and w for this case are both zero. So, since $ΔU+ΔKE=q-w$, we are left with $ΔU=-ΔKE$. The change in kinetic energy of the clay is negative because it is initially moving and then stops. So $ΔU>0$. The positive change in internal energy of the clay is related to the temperature rise by $ΔU=mCΔT$. So the temperature also rises.

Chet