Thermodynamics B/C

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Re: Thermodynamics B/C

Postby UTF-8 U+6211 U+662F » April 12th, 2018, 3:29 pm

Okay, not sure if it's my turn or not but:

Why do different solids have different heat capacities?

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Re: Thermodynamics B/C

Postby CookiePie1 » April 13th, 2018, 8:01 am

Okay, not sure if it's my turn or not but:

Why do different solids have different heat capacities?
Answer: The bonds between the atoms of some materials are stronger, thus needing more heat to raise the temperature
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Re: Thermodynamics B/C

Postby UTF-8 U+6211 U+662F » April 13th, 2018, 1:29 pm

Okay, not sure if it's my turn or not but:

Why do different solids have different heat capacities?
Answer: The bonds between the atoms of some materials are stronger, thus needing more heat to raise the temperature
Yep, your turn.

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Re: Thermodynamics B/C

Postby CookiePie1 » April 13th, 2018, 4:37 pm

How much heat is required to change ice at -20°C to steam at 120°C? (in Joules)

Constants:
Latent heat of fusion of water = 334,000 J/kg°C
Latent heat of vaporization of water = 2258,000 J/kg°C
Specific heat of ice = 2108 J/kg°C
Specific heat of water = 4186 J/kg°C
Specific heat of steam = 1996 J/kg°C
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Re: Thermodynamics B/C

Postby JoeyC » April 13th, 2018, 5:02 pm

I'm sorry but how much does the ice weigh (or what is its mass)?
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Re: Thermodynamics B/C

Postby UTF-8 U+6211 U+662F » April 13th, 2018, 7:20 pm

How much heat is required to change ice at -20°C to steam at 120°C? (in Joules)

Constants:
Latent heat of fusion of water = 334,000 J/kg°C Units are J/kg
Latent heat of vaporization of water = 2258,000 J/kg°C Units are J/kg
Specific heat of ice = 2108 J/kg°C
Specific heat of water = 4186 J/kg°C
Specific heat of steam = 1996 J/kg°C
20*2108*m+334000*m+100*4186*m+2258000*m+20*1996*m = ~3,090,000 J/kg * m = ~3090 J/g * m

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Re: Thermodynamics B/C

Postby UTF-8 U+6211 U+662F » April 15th, 2018, 3:25 pm

I'll just go... Consider a gas with an rms speed of 500 m/s in a box 1 m x 5 m x 1 m at STP. What is its mass?

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Re: Thermodynamics B/C

Postby Justin72835 » April 15th, 2018, 5:01 pm

I'll just go... Consider a gas with an rms speed of 500 m/s in a box 1 m x 5 m x 1 m at STP. What is its mass?
[math]v=\sqrt{\frac{3RT}{M}}[/math]

If you plug in 8.314 for R, 273 for T, and 500 for v, you can solve for M (molar mass) which comes out to be 27.23 g/mol. 

Using the ideal gas law, you can solve for the number of moles in the container, which comes out to be 223.2 mol.

From these two values, we can solve for the mass of the gas, which is [b]6.1 kg[/b].
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Re: Thermodynamics B/C

Postby UTF-8 U+6211 U+662F » April 15th, 2018, 5:06 pm

I'll just go... Consider a gas with an rms speed of 500 m/s in a box 1 m x 5 m x 1 m at STP. What is its mass?
[math]v=\sqrt{\frac{3RT}{M}}[/math]

If you plug in 8.314 for R, 273 for T, and 500 for v, you can solve for M (molar mass) which comes out to be 27.23 g/mol. 

Using the ideal gas law, you can solve for the number of moles in the container, which comes out to be 223.2 mol.

From these two values, we can solve for the mass of the gas, which is [b]6.1 kg[/b].
[math]M * n = \frac{3RT}{v^2} * \frac{PV}{RT} = \frac{3PV}{v^2} = \frac{3 * 10^5 Pa * 5 m^3}{(500 m/s)^2} = 6 kg[/math]
or
[math]m_0 * N = \frac{3kT}{v^2} * \frac{PV}{kT} = \frac{3PV}{v^2} = 6 kg[/math]
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Re: Thermodynamics B/C

Postby MattChina » April 25th, 2018, 2:35 pm

i guess ill restart this.
What is the point at which solid, liquid and gas phases are in equilibrium called?
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