Thermodynamics B/C

UTF-8 U+6211 U+662F
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Re: Thermodynamics B/C

Post by UTF-8 U+6211 U+662F »

Crimesolver wrote:
UTF-8 U+6211 U+662F wrote:
Crimesolver wrote:
1) It is a liquid at 3.3 Celsius 
2)This is because ice has a lower specific heat, which means it melts and heats up more easily.
Reminder to use the
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Ooof totally forgot, was excited that I kinda knew the answer to this one
No worries haha
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Re: Thermodynamics B/C

Post by wec01 »

Crimesolver wrote:
wec01 wrote:Here is a relatively straightforward question:

A 100g ice cube at -10 degrees Celsius is added to 100g of water at 10 degrees Celsius.
1) What is the final state of the solution (phase of matter and temperature)?
2) Why doesn't the solution simply become a mixture of ice and water at 0 degrees Celsius?
1) It is a liquid at 3.3 Celsius 
2)This is because ice has a lower specific heat, which means it melts and heats up more easily.
I realize I miscalculated and used 4.186 kJ/g K instead of J/ g K so I think it actually does end up being an ice water mixture but you got what I was going for with the specific heats.
Your turn!
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Re: Thermodynamics B/C

Post by Crimesolver »

State the caloric theory and why its inaccurate based on our current theory.
Keep on going :)
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Re: Thermodynamics B/C

Post by wec01 »

Crimesolver wrote:State the caloric theory and why its inaccurate based on our current theory.
The caloric theory of heat basically says that heat is a fluid that flows from hot to cold substances, however the current theory, the kinetic theory, says that matter is comprised of molecules that are constantly in motion, temperature is the average kinetic energy of those molecules, and heat increases that energy.
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Re: Thermodynamics B/C

Post by Crimesolver »

wec01 wrote:
Crimesolver wrote:State the caloric theory and why its inaccurate based on our current theory.
The caloric theory of heat basically says that heat is a fluid that flows from hot to cold substances, however the current theory, the kinetic theory, says that matter is comprised of molecules that are constantly in motion, temperature is the average kinetic energy of those molecules, and heat increases that energy.
Your turn!
Keep on going :)
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Re: Thermodynamics B/C

Post by wec01 »

Label the following as isothermal, adiabatic, isochoric, or isobaric:

1. A parcel of air undergoing orographic lifting in the atmosphere.
2. A volume of air in a rigid cylinder being slowly compressed by a piston in a large water bath.
3. A volume of air in a rigid cylinder being quickly compressed by a piston.
4. A volume of air in a rigid cylinder with a piston rising and following freely as the system is heated.
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Re: Thermodynamics B/C

Post by UTF-8 U+6211 U+662F »

wec01 wrote:Label the following as isothermal, adiabatic, isochoric, or isobaric:

1. A parcel of air undergoing orographic lifting in the atmosphere.
2. A volume of air in a rigid cylinder being slowly compressed by a piston in a large water bath.
3. A volume of air in a rigid cylinder being quickly compressed by a piston.
4. A volume of air in a rigid cylinder with a piston rising and following freely as the system is heated.
1. Adiabatic, 2. Isothermal, 3. Adiabatic, 4. Isobaric
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Re: Thermodynamics B/C

Post by wec01 »

UTF-8 U+6211 U+662F wrote:
wec01 wrote:Label the following as isothermal, adiabatic, isochoric, or isobaric:

1. A parcel of air undergoing orographic lifting in the atmosphere.
2. A volume of air in a rigid cylinder being slowly compressed by a piston in a large water bath.
3. A volume of air in a rigid cylinder being quickly compressed by a piston.
4. A volume of air in a rigid cylinder with a piston rising and falling freely as the system is heated.
1. Adiabatic, 2. Isothermal, 3. Adiabatic, 4. Isobaric
Yep, your turn
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Re: Thermodynamics B/C

Post by UTF-8 U+6211 U+662F »

How much work is done on a gas that is isothermally compressed to half its original size at a temperature of 40 degrees Réaumur?
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Re: Thermodynamics B/C

Post by wec01 »

UTF-8 U+6211 U+662F wrote:How much work is done on a gas that is isothermally compressed to half its original size at a temperature of 40 degrees Réaumur?
- nRTln(V[math]_2[/math]/V[math]_1[/math]) = - n * 8.314 J/(mol*K) * 40 °Ré * ln (1/2) = - n * 8.314 J/(mol*K) * 323 K * ln (1/2) = [b]1860 J/mol[/b]
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