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5uper5tring wrote:

Alex-RCHS wrote:

5uper5tring wrote:

Was the 39.88 the highest experimental score (prediction + heat retention + ice bonus) achieved in that event? If you had not added the ice, assuming your prediction was also within 0.5 deg without the ice, what do you think you score would have been?

I don’t know what the highest experimental score was, but for the teams that did not use ice the highest possible score is 40.

Based on my trials we would have gotten about 0.5-1.5 points lower without the ice bonus at this time, temp, and volume (I can’t be sure without knowing he best HRF, but I know it was beneficial). That’s just based on the ice and heat scores. I think our prediction would be made much less accurate without the ice water.

Thank you for the information! Still curious -- why is your prediction less accurate without ice water?

With such low temperatures (and significantly higher volumes) it cools very little, so there is little room for variation. Our temperatures generally drop only 1-5 degrees between our initial measurement and 20-40 minutes of cooling. It's hard to be off by much with numbers like that.

Justin72835 wrote:

JoeyC wrote:HI guys! I just got the BDJH regional test, and looking at the one for C division I just can't figure out how to solve this.
1.)You have a piston filled with 8 moles of a monatomic ideal gas at a temperature of 278 K. It is completely insulated (meaning that there is no heat transfer between the walls of the piston) and it is at rest in a vacuum. The piston has a diameter of 25 cm and has a mass of 4.5 kg. You then apply 50 N of force to the top of the piston.

a. (8 points) Determine the change in temperature of the gas:

Answer:98K Does anyone know what formula they used, and how they used it; Ideal gas law doesn't work with both volume and temperature changing after compression.

Wow, can't believe I missed this haha. I was actually the one who wrote this question. Here's how I intended for the problem to be solved:

Answer (Click to Show Hidden Text)

You don't actually have to solve for internal energy or anything crazy like that. You can use the following adiabatic relationship to solve for everything else:

[math]P_1^{1-\gamma}T_1^\gamma=P_2^{1-\gamma}T_2^\gamma[/math]

P1 can be found by taking the mass of the lid, finding its weight, and dividing it by the area: [b]898.4 Pa[/b]
T1 is given: [b]278 K[/b]
P2 can be found by taking the force applied (50 N), dividing it by the area of the lid, and adding it to P1: [b]1917.0 Pa[/b]

The gas is monatomic which means that the value of gamma is 5/3. Plugging these numbers into the equation listed above and solving for T2 gives: [b]376.4 K[/b]

To solve for the change in temperature, just find the difference between T2 and T1, which gives: 98.4 K or [b]98 K[/b].

Here is the actual test/key for if anyone is interested.

Link to Test
Link to Answers

Feel free to pm me if you have any questions about anything.

Shouldn't you also take atmospheric pressure into consideration when solving this problem??

5uper5tring wrote:

Justin72835 wrote:

JoeyC wrote:HI guys! I just got the BDJH regional test, and looking at the one for C division I just can't figure out how to solve this.
1.)You have a piston filled with 8 moles of a monatomic ideal gas at a temperature of 278 K. It is completely insulated (meaning that there is no heat transfer between the walls of the piston) and it is at rest in a vacuum. The piston has a diameter of 25 cm and has a mass of 4.5 kg. You then apply 50 N of force to the top of the piston.

a. (8 points) Determine the change in temperature of the gas:

Answer:98K Does anyone know what formula they used, and how they used it; Ideal gas law doesn't work with both volume and temperature changing after compression.

Wow, can't believe I missed this haha. I was actually the one who wrote this question. Here's how I intended for the problem to be solved:

Answer (Click to Show Hidden Text)

You don't actually have to solve for internal energy or anything crazy like that. You can use the following adiabatic relationship to solve for everything else:

[math]P_1^{1-\gamma}T_1^\gamma=P_2^{1-\gamma}T_2^\gamma[/math]

P1 can be found by taking the mass of the lid, finding its weight, and dividing it by the area: [b]898.4 Pa[/b]
T1 is given: [b]278 K[/b]
P2 can be found by taking the force applied (50 N), dividing it by the area of the lid, and adding it to P1: [b]1917.0 Pa[/b]

The gas is monatomic which means that the value of gamma is 5/3. Plugging these numbers into the equation listed above and solving for T2 gives: [b]376.4 K[/b]

To solve for the change in temperature, just find the difference between T2 and T1, which gives: 98.4 K or [b]98 K[/b].

Here is the actual test/key for if anyone is interested.

Link to Test
Link to Answers

Feel free to pm me if you have any questions about anything.

Shouldn't you also take atmospheric pressure into consideration when solving this problem??

In the question, I specified that the piston was at rest in a vacuum, meaning that you wouldn't have to worry about atmospheric pressure.

Justin72835 wrote:

5uper5tring wrote:

Justin72835 wrote: Wow, can't believe I missed this haha. I was actually the one who wrote this question. Here's how I intended for the problem to be solved:

Answer (Click to Show Hidden Text)

You don't actually have to solve for internal energy or anything crazy like that. You can use the following adiabatic relationship to solve for everything else:

[math]P_1^{1-\gamma}T_1^\gamma=P_2^{1-\gamma}T_2^\gamma[/math]

P1 can be found by taking the mass of the lid, finding its weight, and dividing it by the area: [b]898.4 Pa[/b]
T1 is given: [b]278 K[/b]
P2 can be found by taking the force applied (50 N), dividing it by the area of the lid, and adding it to P1: [b]1917.0 Pa[/b]

The gas is monatomic which means that the value of gamma is 5/3. Plugging these numbers into the equation listed above and solving for T2 gives: [b]376.4 K[/b]

To solve for the change in temperature, just find the difference between T2 and T1, which gives: 98.4 K or [b]98 K[/b].

Here is the actual test/key for if anyone is interested.

Link to Test
Link to Answers

Feel free to pm me if you have any questions about anything.

Shouldn't you also take atmospheric pressure into consideration when solving this problem??

In the question, I specified that the piston was at rest in a vacuum, meaning that you wouldn't have to worry about atmospheric pressure.

Right. Sorry I didn't read the question carefully

Quick question:
A heat engine absorbs 40J of energy from a hot reservoir and does work while releasing 32J to a cold reservoir. What is the efficiency of this engine?
Is the answer this:
The machine does 8J of work, and releases the remaining 32J as lost heat. Therefore the efficiency is work done/energy in = 8/40 = 20%

Is that right?

Alex-RCHS wrote:Quick question:
A heat engine absorbs 40J of energy from a hot reservoir and does work while releasing 32J to a cold reservoir. What is the efficiency of this engine?
Is the answer this:
The machine does 8J of work, and releases the remaining 32J as lost heat. Therefore the efficiency is work done/energy in = 8/40 = 20%

Is that right?

Yep.

I saw this question in a test on the test exchange and couldn’t figure it out. Can one of you help?
A monatomic, ideal gas at 0C and 1 atm occupies 57.3L. It expands isothermally to 4 times its volume, is compessed isobarically to its original volume, after which it is returned to its original temperature, isochorically. calculate ΔU, q, and w for 1 cycle

Thanks!

The48thYoshi wrote:I saw this question in a test on the test exchange and couldn’t figure it out. Can one of you help?
A monatomic, ideal gas at 0C and 1 atm occupies 57.3L. It expands isothermally to 4 times its volume, is compessed isobarically to its original volume, after which it is returned to its original temperature, isochorically. calculate ΔU, q, and w for 1 cycle

Thanks!

First, an isothermal process means that . Because of this, and due to the ideal gas law, . Work is the integral of pressure over volume, so ln.
, and can be easily calculated to be by the ideal gas law, and is known. Using this, the work done by this process is . This is also equal to the heat inputted during this cycle.

The second process occurs at constant pressure, so the work done is . The pressure at this stage must be one fourth of the original pressure because of the isothermal process, so it is . This yields a work of . The pressure and volume after this process are known, and the change in internal energy can be calculated using . This yields an internal energy change of and the first law of thermodynamics yields a heat value of .

For the final process, and . Thus, the heat inputted is using .

Overall, the ΔU for one process is 0 because U is a state variable. The work is done by the system. The total Q is the sum of all of the other Qs throughout the process, , which makes sense because the total internal energy is 0 for the entire process.

I hope this was helpful even though it was most likely not the shortest path to the solution.

(I also really hope I'm right)

photolithoautotroph wrote:

The48thYoshi wrote:I saw this question in a test on the test exchange and couldn’t figure it out. Can one of you help?
A monatomic, ideal gas at 0C and 1 atm occupies 57.3L. It expands isothermally to 4 times its volume, is compessed isobarically to its original volume, after which it is returned to its original temperature, isochorically. calculate ΔU, q, and w for 1 cycle

Thanks!

First, an isothermal process means that . Because of this, and due to the ideal gas law, . Work is the integral of pressure over volume, so ln.
, and can be easily calculated to be by the ideal gas law, and is known. Using this, the work done by this process is . This is also equal to the heat inputted during this cycle.

The second process occurs at constant pressure, so the work done is . The pressure at this stage must be one fourth of the original pressure because of the isothermal process, so it is . This yields a work of . The pressure and volume after this process are known, and the change in internal energy can be calculated using . This yields an internal energy change of and the first law of thermodynamics yields a heat value of .

For the final process, and . Thus, the heat inputted is using .

Overall, the ΔU for one process is 0 because U is a state variable. The work is done by the system. The total Q is the sum of all of the other Qs throughout the process, , which makes sense because the total internal energy is 0 for the entire process.

I hope this was helpful even though it was most likely not the shortest path to the solution.

(I also really hope I'm right)

Thanks for the help! This actually makes a lot of sense.
The next part of the question asked how ΔU, Q, and W would increase or decrease if it were a diatomic gas. Would you care to explain it to me? Thanks again!

The48thYoshi wrote: The next part of the question asked how ΔU, Q, and W would increase or decrease if it were a diatomic gas. Would you care to explain it to me? Thanks again!

A diatomic gas has more degrees of freedom.

A monoatomic gas has 3 degrees of freedom from motion in x, y, and z directions. Hence U = 3/2 p V (1/2 pV for each degree of freedom). A diatomic gas also has two rotational modes (the one along the axis joining the two atoms doesn't count) and a vibration mode (atoms getting closer together and moving apart).

But the vibration mode isn't excited at normal temperatures, so you'd expect U = 5/2 pV for a diatomic gas.