Thermodynamics B/C

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Re: Thermodynamics B/C

Postby WhatScience? » November 10th, 2017, 8:55 am

platinumfalcon wrote:[hide]
Convection is heat transfer by the mass movement of molecules from one place to another (Giancoli). Convection occurs because hot fluids rise. This is because when a fluid is heated, in most cases, its volume increases. Thus, its density decreases, and it experiences a buoyant force propelling it upwards. [/hide]

Next question: During photosynthesis, cells combine "disorderly" compounds (CO2 and H2O) into "orderly" glucose molecules. Is this a violation of the second law of thermodynamics?


It is not because you can decrease entropy in a certain system while taking heat from another source which will cause entropy of its own. The sun generates light through fusion creating huge amounts of entropy meaning that the total entropy of the UNIVERSE is still increasing, therefore following the second law.
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Re: Thermodynamics B/C

Postby platinumfalcon » November 11th, 2017, 7:50 pm

@above: I'm not so sure about whether the large-scale fusion reactions of the sun directly affect entropy changes involved in photosynthesis reactions, but the correct catch, as you said, was that the entropy of the universe is increasing, even though the entropy of the system appears to decrease. Again, I am unsure of the exact mechanism.

Next question: Describe the anomalous behavior of ice from 0 to 4 degrees Celsius.

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Re: Thermodynamics B/C

Postby WhatScience? » November 11th, 2017, 8:09 pm

platinumfalcon wrote:@above: I'm not so sure about whether the large-scale fusion reactions of the sun directly affect entropy changes involved in photosynthesis reactions, but the correct catch, as you said, was that the entropy of the universe is increasing, even though the entropy of the system appears to decrease. Again, I am unsure of the exact mechanism.

Next question: Describe the anomalous behavior of ice from 0 to 4 degrees Celsius.


The way this works is that the person who answered the question generally gets to ask the next one...unless they take too long.
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Re: Thermodynamics B/C

Postby platinumfalcon » November 12th, 2017, 12:13 pm

Sorry about that. I was just keeping the marathon going, since you hadn't posted a question.

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Re: Thermodynamics B/C

Postby Justin72835 » November 12th, 2017, 3:19 pm

platinumfalcon wrote:@above: I'm not so sure about whether the large-scale fusion reactions of the sun directly affect entropy changes involved in photosynthesis reactions, but the correct catch, as you said, was that the entropy of the universe is increasing, even though the entropy of the system appears to decrease. Again, I am unsure of the exact mechanism.

Next question: Describe the anomalous behavior of ice from 0 to 4 degrees Celsius.


At 4 degrees, water is actually at its highest density. Whether or not you add or remove heat from the water, it will still expand. Because of this, once the water is cooled to its freezing point and it is turned into ice, the ice will actually be less dense than the water around it, which is why ice floats in water and is also why the top of a lake tends to freeze first.

Next question: Consider 3 moles of a diatomic ideal gas in a piston. Suppose the piston is reversibly and isobarically compressed at a pressure of 1073 torr from 15 L to 3 L. Find the change in internal energy and the heat added/removed from the gas.
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Re: Thermodynamics B/C

Postby Alex-RCHS » November 12th, 2017, 3:39 pm

answer
1073 torr = 143055 pascals. Pascal = J/m^3
15L = 15*1000cm^3 = .015m^3
.015m^3 * 143055 J/m^3 = 2146J

3L=.003m^3
.003m^3 * 143055 J/m^3 = 429J

429J-2146J = a loss of 1717J of heat
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Re: Thermodynamics B/C

Postby Justin72835 » November 12th, 2017, 4:37 pm

Alex-RCHS wrote:
answer
1073 torr = 143055 pascals. Pascal = J/m^3
15L = 15*1000cm^3 = .015m^3
.015m^3 * 143055 J/m^3 = 2146J

3L=.003m^3
.003m^3 * 143055 J/m^3 = 429J

429J-2146J = a loss of 1717J of heat


Close. 1717J is actually the work done to the gas. If you want to find the change in internal energy then you can use the formula 5/2(PV). This differs from using 3/2(PV) because it is a diatomic gas, meaning that some of the energy goes into causing the molecules to vibrate rather just its kinetic energy for a monatomic gas. After finding the change in internal energy, you can actually find the heat added or removed from the gas using the First Law.
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Re: Thermodynamics B/C

Postby WhatScience? » November 12th, 2017, 7:17 pm

New question. If you add 16,600 Joules of energy to 1.5 kg of water at 23 C, what will be the final temperature?
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Re: Thermodynamics B/C

Postby Justin72835 » November 12th, 2017, 7:45 pm

WhatScience? wrote:New question. If you add 16,600 Joules of energy to 1.5 kg of water at 23 C, what will be the final temperature?


Answer
16,000 J = 16.6 kJ
Specific heat of water is 4.184 kJ/kg

16.6 = 1.5 * 4.184 * (Tf - 23)

Tf = 25.65 C
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Re: Thermodynamics B/C

Postby Alex-RCHS » November 12th, 2017, 9:26 pm

Justin72835 wrote:
Alex-RCHS wrote:
answer
1073 torr = 143055 pascals. Pascal = J/m^3
15L = 15*1000cm^3 = .015m^3
.015m^3 * 143055 J/m^3 = 2146J

3L=.003m^3
.003m^3 * 143055 J/m^3 = 429J

429J-2146J = a loss of 1717J of heat


Close. 1717J is actually the work done to the gas. If you want to find the change in internal energy then you can use the formula 5/2(PV). This differs from using 3/2(PV) because it is a diatomic gas, meaning that some of the energy goes into causing the molecules to vibrate rather just its kinetic energy for a monatomic gas. After finding the change in internal energy, you can actually find the heat added or removed from the gas using the First Law.

Not to distract from the current question, but can you explain this again? If the change in the energy of the gas is not equal to the work done on the gas, then where does that "extra" work go? (For lack of better phrasing)
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Re: Thermodynamics B/C

Postby Justin72835 » November 12th, 2017, 11:18 pm

Alex-RCHS wrote:
Justin72835 wrote:
Alex-RCHS wrote:
answer
1073 torr = 143055 pascals. Pascal = J/m^3
15L = 15*1000cm^3 = .015m^3
.015m^3 * 143055 J/m^3 = 2146J

3L=.003m^3
.003m^3 * 143055 J/m^3 = 429J

429J-2146J = a loss of 1717J of heat


Close. 1717J is actually the work done to the gas. If you want to find the change in internal energy then you can use the formula 5/2(PV). This differs from using 3/2(PV) because it is a diatomic gas, meaning that some of the energy goes into causing the molecules to vibrate rather just its kinetic energy for a monatomic gas. After finding the change in internal energy, you can actually find the heat added or removed from the gas using the First Law.

Not to distract from the current question, but can you explain this again? If the change in the energy of the gas is not equal to the work done on the gas, then where does that "extra" work go? (For lack of better phrasing)


Good question.

You're asking why the work you do on the gas doesn't contribute completely to a change in internal energy. Try to look at it this way. If you actually had the piston in front of you and you tried compressing the gas by pushing down on the top of it, you would feel the pressure increase. In fact, the more you try to compress the gas, the more pressure you'll feel the gas apply back. This is different from the scenario I stated before because the pressure is changing with the volume.

Now imagine this scenario: You have the same piston, but you put ice around it to cool the gas. You can see the top of the piston lower naturally (see Charles' Law). What's happening in this situation is that the ice is drawing heat away from the piston (heat removed), which contributes to lowering the temperature of the gas and making its volume shrink in size. It is clear here that the pressure inside the piston remains constant, as it is still supporting the weight of the piston as well as the atmospheric pressure. This also means that the surroundings are doing work on the gas.

The second scenario is the one laid out in the problem I gave. It is clear here that the surroundings are doing work on the gas, but since it is at a constant pressure, the gas is actually losing heat as well. This all results in a decrease in internal energy.

Here's the complete answer for those who are wondering
The First Law of Thermodynamics can be written as . We already know that the work is being done on the gas since it is being compressed. The next step at this point is to find the change in internal energy of the gas.

Since it is a diatomic gas, the equation can be applied here. We know that from the Ideal Gas Law, so we can substitute this term into the previous equation to get . After solving this out, we find that change in internal energy is -4292 J.

Plugging in this value into the first equation, we can solve for the heat added/removed, which turns out to be -6008 J (heat is removed).

Change in internal energy is -4291 J
Heat removed is 6008 J
Work done on gas is 1717 J
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Re: Thermodynamics B/C

Postby WhatScience? » November 13th, 2017, 5:33 pm

What is the equation for rate of thermal conduction, what do the factors in it mean, and why do the different factors make sense?
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Re: Thermodynamics B/C

Postby Justin72835 » November 13th, 2017, 8:12 pm

WhatScience? wrote:What is the equation for rate of thermal conduction, what do the factors in it mean, and why do the different factors make sense?




The term on the left side of the equation describes the rate of heat transfer through conduction.

'k' is the thermal conductivity. This makes sense because the rate of conduction obviously depends on the material used (metal vs aerogel).
'A' is the area. This makes senses because the number of molecular collisions occurring increases with a greater cross-sectional area. Think about how touching a hot object with your finger differs from touching the same object with the palm of your hand.
'L' is the length. This makes sense because thermal energy will take longer to travel through a thicker object. Think about a thin piece of paper versus a block of wood.
'T1' and 'T2' are temperatures of the surroundings and the object, respectively. This makes sense because with a higher difference in temperatures there is a larger temperature gradient. Think of temperature gradient as the slope of a mountain. If your slope is really steep (large gradient), then objects will roll down the mountain in a much shorter time (increased rate of heat transfer), and vice versa for a mountain with little to no slope (no difference in temperature, already at thermal equilibrium).
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Re: Thermodynamics B/C

Postby Justin72835 » November 13th, 2017, 8:13 pm

When you get out of a pool, even the slightest breeze will make you feel cold. Why is this?
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Re: Thermodynamics B/C

Postby Crtomir » December 2nd, 2017, 7:18 am

Justin72835 wrote:When you get out of a pool, even the slightest breeze will make you feel cold. Why is this?


Answer: Evaporation of water is an endothermic process, meaning it sucks in heat from the environment, which includes your skin which is at a higher temperature than the water. Since it sucks heat energy form your skin, you feel cold.

Next Question: (From Shortley and Williams Elements of Physics, 3rd Edition) A layer of ice on a pond is 2 cm thick. When the upper surface of the ice is at -15C and the temperature of the water just below the ice is 0C, at what rate does the ice become thicker? Express your answer in units of [cm/min.]. Show your work.


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