Physics Help Forum Determine the increase in entropy

 Thermodynamics and Fluid Mechanics Thermodynamics and Fluid Mechanics Physics Help Forum

 Feb 7th 2018, 08:17 AM #1 Junior Member   Join Date: Feb 2018 Posts: 9 Determine the increase in entropy A 60 kg mass is dragged horizontally over a distance of 10 m across the floor. The friction coefficient between the mass and the floor is 0.2. Determine the entropy increase of the surrounding atmosphere which is at 25 °C.
 Feb 7th 2018, 08:43 AM #2 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 926 Can you explain in words - qualitatively - why you think there is an increase in entropy in the atmosphere? Once you understand what is going on you can tackle the numbers.
 Feb 7th 2018, 09:21 AM #3 Junior Member   Join Date: Feb 2018 Posts: 9 Based on what i know, the friction produced by the 60kg mass will generate additional entropy that will go into the surrounding
Feb 7th 2018, 09:38 AM   #4
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 Originally Posted by casperneo Based on what i know, the friction produced by the 60kg mass will generate additional entropy that will go into the surrounding
Yes so how (in what form) is that entropy transferred to the atmosphere?

 Feb 7th 2018, 06:48 PM #5 Junior Member   Join Date: Feb 2018 Posts: 9 By kinetic energry?
 Feb 8th 2018, 12:34 AM #6 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 926 Does the mass start from rest and end at rest? So overall, does it retain any kinetic energy after being dragged? What sort of energy appears when you rub two things together (friction)? So what happens to all the work done in dragging the mass? That is the application of the first law. The atmosphere is a very large body is this amount of energy going to change its temperature? What is the definition of entropy? That is the application of the second law, which give you your final answer, if you follow these steps.
 Feb 9th 2018, 04:50 AM #7 Junior Member   Join Date: Feb 2018 Posts: 9 using the friction force of 117.72N, i found that work done by dragging the mass is 1177.2Nm. Applying first law, work done by dragging = change in internal energy -1177.2 = changing in internal energy how should i proceed?
 Feb 9th 2018, 05:52 AM #8 Senior Member   Join Date: Oct 2017 Location: Glasgow Posts: 167 Casperneo, the basic definition of the entropy is $\displaystyle S = \frac{\Delta Q}{T}$ where $\displaystyle \Delta Q$ is the change in energy (a heat transfer) and $\displaystyle T$ is temperature, using SI units.
Feb 9th 2018, 06:28 AM   #9
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 Originally Posted by casperneo using the friction force of 117.72N, i found that work done by dragging the mass is 1177.2Nm. Applying first law, work done by dragging = change in internal energy -1177.2 = changing in internal energy how should i proceed?
Was there a change in internal energy?
If so the internal energy of what?

I think the change of internal energy was zero.

That is because the work against friction generates heat.

This heat is transferred to the atmosphere (you have to assume the whole of it is transferred and that none goes into heating the ground or the block or you can't do the question. Really the question should have said "heat transferred to the surroundings.")

So the sum of the work done plus the heat transferred equals zero.

Therefore heat transferred = work done

That is the proper application of the first law.

Then you can proceed to the next step

Which was "does this small amount of heat change the temperature of the atmosphere appreciably?"

The answer to that allows you to calculate the entropy change in the atmosphere following Benit's formula.

If that heat had warmed the air appreciably could you still have used this?

 Tags determine, entropy, increase

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