Thermodynamics B/C

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WhatScience?
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Re: Thermodynamics B/C

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

[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.
"When you clean your room, you are increasing the total chaos of the universe" - Hank Green Crash Course (Entropy)

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Thermodynamics, Potions and Poisons, Disease Detectives, Optics, and Towers

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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

@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.
"When you clean your room, you are increasing the total chaos of the universe" - Hank Green Crash Course (Entropy)

Events 2018

Thermodynamics, Potions and Poisons, Disease Detectives, Optics, and Towers

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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

@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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Alex-RCHS
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Re: Thermodynamics B/C

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

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

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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Events 2018

Thermodynamics, Potions and Poisons, Disease Detectives, Optics, and Towers

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Re: Thermodynamics B/C

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

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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