## Thermodynamics B/C

Justin72835
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### Re: Thermodynamics B/C

Alex-RCHS wrote:
Justin72835 wrote:
Alex-RCHS wrote:
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.
The First Law of Thermodynamics can be written as $\Delta U = Q + W$. 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 $\Delta U = \frac {5}{2} nR \Delta T$ can be applied here. We know that $PV = nRT$ from the Ideal Gas Law, so we can substitute this term into the previous equation to get $\Delta U = \frac {5}{2} P \Delta V$. 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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WhatScience?
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### Re: Thermodynamics B/C

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

Justin72835
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### Re: Thermodynamics B/C

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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Justin72835
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### Re: Thermodynamics B/C

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

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.

Justin72835
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### Re: Thermodynamics B/C

Crtomir wrote:
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.
$\frac{dQ}{dt} = k\frac{A(T_h - T_c)}{L}$

Plugging in the numbers given and knowing that the thermal conductivity of ice is approximately 2.25 W/mk:

$\frac{dQ}{dt} = 2.25\frac{0.0001(0 - (-15))}{0.02} = 0.16875 W$

These means that 0.16875 Joules are being transferred across each cm^2 of ice every second. We know that the the latent heat of fusion is 334 Joules/gram and that the density of ice is 0.92 g/mL. Now we can find the mass and volume of ice formed:

$\frac{dM}{dt} = \frac{dQ}{dt} * \frac{1}{L_{fusion}} = 0.000505 g/s$

$\frac{dV}{dt} = \frac{dM}{dt} * \frac{1}{\rho} = 0.000549 cm^3/s$

The area is 1 cm^2 so we can plug that into the next equation. To find the rate of ice formation:

$\frac{dL}{dt} = \frac{dV}{dt} * \frac{1}{dA} = 0.000549 cm/s$

This answer is in centimeters per second, so multiply by 60 to find the number of centimeters of ice formed per minute:

[b]0.03295 cm/min[/b]
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Crtomir
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### Re: Thermodynamics B/C

Justin72835 wrote:
Crtomir wrote:
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.
$\frac{dQ}{dt} = k\frac{A(T_h - T_c)}{L}$

Plugging in the numbers given and knowing that the thermal conductivity of ice is approximately 2.25 W/mk:

$\frac{dQ}{dt} = 2.25\frac{0.0001(0 - (-15))}{0.02} = 0.16875 W$

These means that 0.16875 Joules are being transferred across each cm^2 of ice every second. We know that the the latent heat of fusion is 334 Joules/gram and that the density of ice is 0.92 g/mL. Now we can find the mass and volume of ice formed:

$\frac{dM}{dt} = \frac{dQ}{dt} * \frac{1}{L_{fusion}} = 0.000505 g/s$

$\frac{dV}{dt} = \frac{dM}{dt} * \frac{1}{\rho} = 0.000549 cm^3/s$

The area is 1 cm^2 so we can plug that into the next equation. To find the rate of ice formation:

$\frac{dL}{dt} = \frac{dV}{dt} * \frac{1}{dA} = 0.000549 cm/s$

This answer is in centimeters per second, so multiply by 60 to find the number of centimeters of ice formed per minute:

[b]0.03295 cm/min[/b]

Correct. Nice Job!

Justin72835
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### Re: Thermodynamics B/C

Great! Now it's my turn.

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.

Determine the change in temperature of the gas.
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### Re: Thermodynamics B/C

Justin72835 wrote:Great! Now it's my turn.

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.

Determine the change in temperature of the gas.
P = F/A = (4.5 kg * 9.8 m/s^2)/(0.0625pi m^2) = 224.599 Pa
V = nRT/P = (8 mol)R(278 K)/224.599 Pa =
82.331 m^3
P2 = F/A = (4.5 kg * 9.8 m/s^2 + 50 N)/(0.0625pi m^2) = 479.247 Pa
Since it's adiabatic compression on a monatomic ideal gas, PV^(5/3) = (P2)(V2)^(5/3);
V2 = (PV^(5/3)/P2)^(3/5) = 52.248 m^3
PV/T = (P2)(V2)/(T2); T2 = (P2)(V2)T/(PV)
= 376.448 K
T2 - T = [b]98 K[/b]

Justin72835
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### Re: Thermodynamics B/C

UTF-8 U+6211 U+662F wrote:
Justin72835 wrote:Great! Now it's my turn.

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.

Determine the change in temperature of the gas.
P = F/A = (4.5 kg * 9.8 m/s^2)/(0.0625pi m^2) = 224.599 Pa
V = nRT/P = (8 mol)R(278 K)/224.599 Pa =
82.331 m^3
P2 = F/A = (4.5 kg * 9.8 m/s^2 + 50 N)/(0.0625pi m^2) = 479.247 Pa
Since it's adiabatic compression on a monatomic ideal gas, PV^(5/3) = (P2)(V2)^(5/3);
V2 = (PV^(5/3)/P2)^(3/5) = 52.248 m^3
PV/T = (P2)(V2)/(T2); T2 = (P2)(V2)T/(PV)
= 376.448 K
T2 - T = [b]98 K[/b]
You got it right, nice job!

One thing I noticed while looking through your work is that when you were calculating for pressure and area, you used the diameter instead of the radius of the piston. This didn't really matter in the end since the pressures got divided out in the final formula (causing the areas to cancel out), but it's something that you should definitely watch for.
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