Thermodynamics B/C

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Re: Thermodynamics B/C

Postby Justin72835 » February 12th, 2018, 6:39 am

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Sorry to interrupt your guys's chain of questions but I can finally answer one! Since we are dealing with an ideal gas, we can use the ideal gas law which states PV = nRT. Volume and R stay constant, so we can ignore those. The initial ratio of P/nT must equal the final ratio of P/nT. Lets assume we start with a pressure of 1 to keep the math simple. This gives us an equation 1/(46*(273+22)) = .5/(n*(273+22-13)). Solving for n gives us a final mass of 24 kg for the air. The question asks for how much is removed, so subtracting 24.1 from 46 gives us 22 kg of air pumped out of the room.
Way to ruin the fun :roll:

Jk haha, that's the right answer. Go ahead with the next question!
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Re: Thermodynamics B/C

Postby Riptide » February 12th, 2018, 1:55 pm

There is a uniform sphere that can be approximated as being a blackbody. The sphere has a radius of 20 cm and has a temperature of 4300 K. The temperature of the surroundings is 300 K.

Find the rate of heat transfer between the sphere and the surroundings.


If you wanted to halve the rate of heat transfer, then what should you change the temperature of the sphere to?
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Re: Thermodynamics B/C

Postby Justin72835 » February 12th, 2018, 2:51 pm

There is a uniform sphere that can be approximated as being a blackbody. The sphere has a radius of 20 cm and has a temperature of 4300 K. The temperature of the surroundings is 300 K.

Find the rate of heat transfer between the sphere and the surroundings.

If you wanted to halve the rate of heat transfer, then what should you change the temperature of the sphere to?
Answer
Because this problem deals with radiation, you have to use the Stefan-Boltzmann Law. [math]\frac{dQ}{dt}=A\sigma (T_1^4-T_2^4)=4\pi r^2\sigma (T_1^4-T_2^4)=9.74 * 10^6\ W[/math] To find the temperature when the rate of heat transfer is halved, set dQ/dT to one-half of the previous answer and solve for T1. which gives an answer of 3616K. Final Answers: 9.74 * 10^6W, 3616K
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Re: Thermodynamics B/C

Postby Alex-RCHS » February 12th, 2018, 2:58 pm

There is a uniform sphere that can be approximated as being a blackbody. The sphere has a radius of 20 cm and has a temperature of 4300 K. The temperature of the surroundings is 300 K.

Find the rate of heat transfer between the sphere and the surroundings.


If you wanted to halve the rate of heat transfer, then what should you change the temperature of the sphere to?
This feels more like Remote to me, but
Answer
I mean technically a black body is in thermodynamic equilibrium with its surroundings, but I’m guess that’s not the answer you want. The rate of radiative heat emission per unit surface area (radiative flux is the proper name I think) is given by the Stefan-Boltzmann law. By multiplying by the surface area, you get: W=(emissivity)(5.67*10^-8)(Temperature^4)(surface area) W=(1)(5.67*10^-8)(3.42*10^14)(.503) W=9.75*10^6 watts That’s the rate of heat transfer via black dot radiation out of the sphere, and also the rate at which heat is absorbed by the sphere from its surroundings. But like I said at the beginning, the rates are equal (hence the thermodynamic equilibrium clause in the definition of a black body), so technically the answer is zero.
Edit: beat me to it. Please note my point about the phrasing of the question, though.
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Re: Thermodynamics B/C

Postby Riptide » February 12th, 2018, 3:17 pm

There is a uniform sphere that can be approximated as being a blackbody. The sphere has a radius of 20 cm and has a temperature of 4300 K. The temperature of the surroundings is 300 K.

Find the rate of heat transfer between the sphere and the surroundings.


If you wanted to halve the rate of heat transfer, then what should you change the temperature of the sphere to?
This feels more like Remote to me, but
Answer
I mean technically a black body is in thermodynamic equilibrium with its surroundings, but I’m guess that’s not the answer you want. The rate of radiative heat emission per unit surface area (radiative flux is the proper name I think) is given by the Stefan-Boltzmann law. By multiplying by the surface area, you get: W=(emissivity)(5.67*10^-8)(Temperature^4)(surface area) W=(1)(5.67*10^-8)(3.42*10^14)(.503) W=9.75*10^6 watts That’s the rate of heat transfer via black dot radiation out of the sphere, and also the rate at which heat is absorbed by the sphere from its surroundings. But like I said at the beginning, the rates are equal (hence the thermodynamic equilibrium clause in the definition of a black body), so technically the answer is zero.
Edit: beat me to it. Please note my point about the phrasing of the question, though.
Correct me if I'm wrong, but how can the two objects be in thermodynamic equilibrium when they are at different temperatures? For example, lets say that a hot piece of coal (which is we can approximate as a blackbody) was left out at room temperature, it will continuously cool down until it reaches thermodynamic equilibrium with its surroundings.

PS - A friend that does remsen actually helped me write this, since I don't actually do this event, so sorry if it's a bit off topic xD
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Re: Thermodynamics B/C

Postby Alex-RCHS » February 12th, 2018, 3:42 pm

Correct me if I'm wrong, but how can the two objects be in thermodynamic equilibrium when they are at different temperatures? For example, lets say that a hot piece of coal (which is we can approximate as a blackbody) was left out at room temperature, it will continuously cool down until it reaches thermodynamic equilibrium with its surroundings.
Well, first of all, I was wrong about what I said earlier, my apologies. I think that one requirement for black body *radiation* to occur is that the black body is in thermal equilibrium with its surroundings, but I don’t think that’s a requirement for something to simply *be* a black body. But I could be wrong about that.

Anyways, since you said to approximate it as a black body then there’s no reason to assume that the object is in thermodynamic equilibrium with its surroundings, so that’s my bad.

Also, they could be in thermodynamic equilibrium at different temperatures because of radiation, but in this case that wouldn’t occur because the hotter object has a higher emissivity (and because the difference in temperatures wasn’t so large).
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Re: Thermodynamics B/C

Postby Riptide » February 12th, 2018, 6:44 pm

There is a uniform sphere that can be approximated as being a blackbody. The sphere has a radius of 20 cm and has a temperature of 4300 K. The temperature of the surroundings is 300 K.

Find the rate of heat transfer between the sphere and the surroundings.

If you wanted to halve the rate of heat transfer, then what should you change the temperature of the sphere to?
Answer
Because this problem deals with radiation, you have to use the Stefan-Boltzmann Law. [math]\frac{dQ}{dt}=A\sigma (T_1^4-T_2^4)=4\pi r^2\sigma (T_1^4-T_2^4)=9.74 * 10^6\ W[/math] To find the temperature when the rate of heat transfer is halved, set dQ/dT to one-half of the previous answer and solve for T1. which gives an answer of 3616K. Final Answers: 9.74 * 10^6W, 3616K
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Re: Thermodynamics B/C

Postby Justin72835 » February 12th, 2018, 6:57 pm

There is a uniform sphere that can be approximated as being a blackbody. The sphere has a radius of 20 cm and has a temperature of 4300 K. The temperature of the surroundings is 300 K.

Find the rate of heat transfer between the sphere and the surroundings.

If you wanted to halve the rate of heat transfer, then what should you change the temperature of the sphere to?
Answer
Because this problem deals with radiation, you have to use the Stefan-Boltzmann Law. [math]\frac{dQ}{dt}=A\sigma (T_1^4-T_2^4)=4\pi r^2\sigma (T_1^4-T_2^4)=9.74 * 10^6\ W[/math] To find the temperature when the rate of heat transfer is halved, set dQ/dT to one-half of the previous answer and solve for T1. which gives an answer of 3616K. Final Answers: 9.74 * 10^6W, 3616K
Nice job! Your turn
A pressure gauge indicates the differences between atmospheric pressure and pressure inside the tank. The gauge on a 2.00 m^3 oxygen tank reads 36 atm. After some use of the oxygen, the gauge reads 24 atm. How many cubic meters of oxygen at normal atmospheric pressure was used? There is no temperature change during the time of consumption.
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Re: Thermodynamics B/C

Postby UTF-8 U+6211 U+662F » February 12th, 2018, 7:31 pm

A pressure gauge indicates the differences between atmospheric pressure and pressure inside the tank. The gauge on a 2.00 m^3 oxygen tank reads 36 atm. After some use of the oxygen, the gauge reads 24 atm. How many cubic meters of oxygen at normal atmospheric pressure was used? There is no temperature change during the time of consumption.
Answer
Boyle's law: PV is constant. (37 atm)(2.00 m^3) = (25 atm)V. V = (37/25 * 2.00 m^3) = 2.96 m^3 = 3.0 m^3.

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Re: Thermodynamics B/C

Postby Justin72835 » February 12th, 2018, 7:42 pm

A pressure gauge indicates the differences between atmospheric pressure and pressure inside the tank. The gauge on a 2.00 m^3 oxygen tank reads 36 atm. After some use of the oxygen, the gauge reads 24 atm. How many cubic meters of oxygen at normal atmospheric pressure was used? There is no temperature change during the time of consumption.
Answer
Boyle's law: PV is constant. (37 atm)(2.00 m^3) = (25 atm)V. V = (37/25 * 2.00 m^3) = 2.96 m^3 = 3.0 m^3.
Not quite ;)
Answer
Perhaps I didn't make it very clear but what is happening in this question is that some oxygen is released from the tank and you are trying to find the volume of that released air at 1 atm. Your first step was correct; the volume of [i]all[/i] the air at a pressure of 25 atm would be 2.96 m^3. 2.00 m^3 of this would stay in the tank, while the other 0.96 m^3 would be released out into the atmosphere. Because this is the volume of the released air at 25 atm, you need to use Boyle's Law once again to find its volume at 1 atm. (0.96 m^3)(25 atm) = (1 atm)(V), from which we find that V = 24 m^3. You can go ahead with the next question though :D
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Re: Thermodynamics B/C

Postby UTF-8 U+6211 U+662F » February 13th, 2018, 1:58 pm

A pressure gauge indicates the differences between atmospheric pressure and pressure inside the tank. The gauge on a 2.00 m^3 oxygen tank reads 36 atm. After some use of the oxygen, the gauge reads 24 atm. How many cubic meters of oxygen at normal atmospheric pressure was used? There is no temperature change during the time of consumption.
Answer
Boyle's law: PV is constant. (37 atm)(2.00 m^3) = (25 atm)V. V = (37/25 * 2.00 m^3) = 2.96 m^3 = 3.0 m^3.
Not quite ;)
Answer
Perhaps I didn't make it very clear but what is happening in this question is that some oxygen is released from the tank and you are trying to find the volume of that released air at 1 atm. Your first step was correct; the volume of [i]all[/i] the air at a pressure of 25 atm would be 2.96 m^3. 2.00 m^3 of this would stay in the tank, while the other 0.96 m^3 would be released out into the atmosphere. Because this is the volume of the released air at 25 atm, you need to use Boyle's Law once again to find its volume at 1 atm. (0.96 m^3)(25 atm) = (1 atm)(V), from which we find that V = 24 m^3. You can go ahead with the next question though :D
Oh, whoops. Next question: If a Carnot engine operates at 50 degrees Fahrenheit and 80 degrees Fahrenheit, find the change in entropy after one cycle. What would be the change in entropy if the Carnot engine were a Carnot refrigerator? Explain why the Carnot engine is the most efficient heat engine between two temperature sources.

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Re: Thermodynamics B/C

Postby Justin72835 » February 13th, 2018, 5:19 pm

Answer
Boyle's law: PV is constant. (37 atm)(2.00 m^3) = (25 atm)V. V = (37/25 * 2.00 m^3) = 2.96 m^3 = 3.0 m^3.
Not quite ;)
Answer
Perhaps I didn't make it very clear but what is happening in this question is that some oxygen is released from the tank and you are trying to find the volume of that released air at 1 atm. Your first step was correct; the volume of [i]all[/i] the air at a pressure of 25 atm would be 2.96 m^3. 2.00 m^3 of this would stay in the tank, while the other 0.96 m^3 would be released out into the atmosphere. Because this is the volume of the released air at 25 atm, you need to use Boyle's Law once again to find its volume at 1 atm. (0.96 m^3)(25 atm) = (1 atm)(V), from which we find that V = 24 m^3. You can go ahead with the next question though :D
Oh, whoops. Next question: If a Carnot engine operates at 50 degrees Fahrenheit and 80 degrees Fahrenheit, find the change in entropy after one cycle. What would be the change in entropy if the Carnot engine were a Carnot refrigerator? Explain why the Carnot engine is the most efficient heat engine between two temperature sources.
Answer
Entropy is a state function, meaning that it depends only on its initial and final conditions. Because of this, after one complete cycle, the change in entropy for the system is 0 J/K. This is true for both the Carnot engine and the Carnot refrigerator. For the last part of the question, the Carnot engine is the most efficient between two temperature sources because all of its steps are reversible.
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Re: Thermodynamics B/C

Postby UTF-8 U+6211 U+662F » February 13th, 2018, 5:21 pm

Not quite ;)
Answer
Perhaps I didn't make it very clear but what is happening in this question is that some oxygen is released from the tank and you are trying to find the volume of that released air at 1 atm. Your first step was correct; the volume of [i]all[/i] the air at a pressure of 25 atm would be 2.96 m^3. 2.00 m^3 of this would stay in the tank, while the other 0.96 m^3 would be released out into the atmosphere. Because this is the volume of the released air at 25 atm, you need to use Boyle's Law once again to find its volume at 1 atm. (0.96 m^3)(25 atm) = (1 atm)(V), from which we find that V = 24 m^3. You can go ahead with the next question though :D
Oh, whoops. Next question: If a Carnot engine operates at 50 degrees Fahrenheit and 80 degrees Fahrenheit, find the change in entropy after one cycle. What would be the change in entropy if the Carnot engine were a Carnot refrigerator? Explain why the Carnot engine is the most efficient heat engine between two temperature sources.
Answer
Entropy is a state function, meaning that it depends only on its initial and final conditions. Because of this, after one complete cycle, the change in entropy for the system is 0 J/K. This is true for both the Carnot engine and the Carnot refrigerator. For the last part of the question, the Carnot engine is the most efficient between two temperature sources because all of its steps are reversible.
Yep! Although you could explain the last part a little more (non sequitur there). Your turn!

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Re: Thermodynamics B/C

Postby Justin72835 » February 13th, 2018, 6:00 pm

Oh, whoops. Next question: If a Carnot engine operates at 50 degrees Fahrenheit and 80 degrees Fahrenheit, find the change in entropy after one cycle. What would be the change in entropy if the Carnot engine were a Carnot refrigerator? Explain why the Carnot engine is the most efficient heat engine between two temperature sources.
Answer
Entropy is a state function, meaning that it depends only on its initial and final conditions. Because of this, after one complete cycle, the change in entropy for the system is 0 J/K. This is true for both the Carnot engine and the Carnot refrigerator. For the last part of the question, the Carnot engine is the most efficient between two temperature sources because all of its steps are reversible.
Yep! Although you could explain the last part a little more (non sequitur there). Your turn!
Alright. A cup of tea cools from 170 degrees Fahrenheit to 145 degrees Fahrenheit in 6 minutes in a room at 72 degrees Fahrenheit. How long will it take for the tea to cool from 100 to 90 degrees Fahrenheit in the same room?
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Re: Thermodynamics B/C

Postby Riptide » February 14th, 2018, 6:07 pm

Alright. A cup of tea cools from 170 degrees Fahrenheit to 145 degrees Fahrenheit in 6 minutes in a room at 72 degrees Fahrenheit. How long will it take for the tea to cool from 100 to 90 degrees Fahrenheit in the same room?
Answer
Using Newton's Law of Cooling, we get T = Ts + (T0-Ts)e^kt. Plugging in our data gives us a k value of -4.91E-2. We can set up this equation again with 100 to 90 degrees, and calculate a t of 9.0 minutes.
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