Entropy change of a hot, falling object

In summary, a shipyard worker drops a hot steel rivet (mass 125g, temperature 350 degrees C)into a river at temperature 5 degrees C, a distance 30m below. The entropy change of the rivet is 17250 J, and the entropy change of the river is 62.1 JK-1. The net increase in entropy for the "universe" is 21.93 J/K.
  • #1
freddy_12345
3
0

Homework Statement



A shipyard worker drops a hot steel rivet (mass 125g, temperature 350 degrees C)
into a river at temperature 5 degrees C, a distance 30m below. Stating any assumptions
you make, calculate the entropy change of the universe as a result of this event.
(specific heat capacity of steel ~0.4 J g-1 K-1).

Homework Equations


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The Attempt at a Solution



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If anybody could tell me if I'm on the right lines at all or what the heck to do with the fact that the rivet's being dropped from 30m I'd greatly appreciate it. Thanks!
 
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  • #2
You have traced the change in entropy of the rivet, but I believe there are two more things you need to look at in the "universe" (essentially the rivet-Earth system). How much heat has been added to the river by the change in gravitational potential energy in the rivet-Earth system? Also, the river absorbs the heat of the rivet with virtually no temperature change (it's large enough to serve as a "heat reservoir"), so what is the change in the entropy of the river from that?

So there is the entropy change of the cooling rivet plus the entropy change from the change in mechanical energy plus the entropy change from the heat transfer from rivet to river.
 
  • #3
Thanks for the reply.

So, as I see it, so far I've worked out the decrease in the entropy of the rivet.

The heat flowing into the reservoir from the rivet cooling will be the same as the heat loss from the rivet:

1.jpg


That gives me 17250 J.

Then, if we assume the temperature of the reservoir doesn't change, then, using the temperature of the river in Kelvins for T:

7.jpg


That gives me an increase of entropy in the river as 62.1 JK-1

Finally, the change in gravitational potential energy is:

8.jpg


If we assume that this is all converted to heat in the river we can work out a second component for the increase in entropy. I get S = 0.13 JK-1.


Overall this gives me a net increase in the entropy of the universe as S = 21.93 JK-1.

Is this the right track or have I completely messed up?
 
  • #4
I agree with those results to the first decimal place (since I rounded off only at the end), so we get a net increase in entropy for the "universe" of 21.9 J/K .

It is a reasonable safe assumption that just about all of the potential energy change is transferred into heating the river. The "splash" at impact displaces water and ejects some of it upward (briefly), as well as producing (a minute amount of) acoustical energy. But since practically all the water falls back into the river, that mechanical energy winds up heating the river after all. (We can extend "the universe" to include the air over the river, but the tiny amounts of water scattered as microdroplets and vapor and of sound energy probably contribute additions to entropy scarcely worth pursuing...)
 
  • #5


I would approach this problem by first making a few assumptions. I would assume that the steel rivet is in thermal equilibrium with its surroundings before it is dropped, and that the river is a large enough body of water that its temperature remains constant during the experiment. I would also assume that there is no significant energy loss due to air resistance as the rivet falls.

Using these assumptions, we can calculate the initial entropy of the universe, which is given by the equation S = k ln(W), where k is the Boltzmann constant and W is the number of microstates available to the system. Since the rivet is in thermal equilibrium with its surroundings before it is dropped, the number of microstates is equal to 1, so the initial entropy of the universe is 0.

Next, we can calculate the final entropy of the universe after the rivet has fallen into the river. We know that the specific heat capacity of steel is 0.4 J g-1 K-1, so we can use the equation Q = mcΔT to calculate the change in internal energy of the rivet as it cools down. We know the mass of the rivet (125g), the specific heat capacity (0.4 J g-1 K-1), and the change in temperature (350 degrees C - 5 degrees C = 345 degrees C). This gives us a change in internal energy of Q = (125g)(0.4 J g-1 K-1)(345 degrees C) = 17250 J.

We can then use the equation ΔS = Q/T to calculate the change in entropy of the rivet. Since the rivet is losing energy, its entropy decreases, so ΔS = -17250 J / 298 K (I have assumed a standard temperature of 298 K for the river water). This gives us a change in entropy of -58 J/K.

Finally, we can calculate the change in entropy of the universe by adding the initial entropy (0) to the change in entropy of the rivet (-58 J/K). This gives us a final entropy of -58 J/K for the universe after the rivet has fallen into the river.

In summary, the entropy of the universe decreases by -58 J/K as a result of the hot, falling object. This is due to the fact that the rivet loses energy and its entropy decreases as it cools down, while the
 

What is entropy change?

Entropy change refers to the measure of disorder or randomness in a system. It is a thermodynamic property that describes the distribution of energy within a system.

How does the temperature of a falling object affect its entropy change?

As a hot object falls, its temperature decreases. This decrease in temperature causes a decrease in entropy, as the molecules of the object become more ordered and have less random motion.

What factors can influence the entropy change of a falling object?

The mass, speed, and initial temperature of the object are all factors that can influence the entropy change. Additionally, external factors such as air resistance and friction can also play a role.

Can the entropy change of a falling object be negative?

Yes, it is possible for the entropy change of a falling object to be negative. This can occur if the object is falling in a highly controlled environment, such as a vacuum, where there is minimal external influence on the object's motion.

How is the concept of entropy change related to the second law of thermodynamics?

The second law of thermodynamics states that the total entropy of a closed system will always increase or remain constant over time. As a falling object loses energy and its temperature decreases, the overall entropy of the system increases in accordance with this law.

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