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Re: Thermodynamics B/C

Posted: April 12th, 2018, 3:29 pm
by UTF-8 U+6211 U+662F
Okay, not sure if it's my turn or not but:

Why do different solids have different heat capacities?

Re: Thermodynamics B/C

Posted: April 13th, 2018, 8:01 am
by CookiePie1
UTF-8 U+6211 U+662F wrote: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

Re: Thermodynamics B/C

Posted: April 13th, 2018, 1:29 pm
by UTF-8 U+6211 U+662F
CookiePie1 wrote:
UTF-8 U+6211 U+662F wrote: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.

Re: Thermodynamics B/C

Posted: April 13th, 2018, 4:37 pm
by CookiePie1
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

Re: Thermodynamics B/C

Posted: April 13th, 2018, 5:02 pm
by JoeyC
I'm sorry but how much does the ice weigh (or what is its mass)?

Re: Thermodynamics B/C

Posted: April 13th, 2018, 7:20 pm
by UTF-8 U+6211 U+662F
CookiePie1 wrote: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

Re: Thermodynamics B/C

Posted: April 15th, 2018, 3:25 pm
by UTF-8 U+6211 U+662F
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?

Re: Thermodynamics B/C

Posted: April 15th, 2018, 5:01 pm
by Justin72835
UTF-8 U+6211 U+662F wrote: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].

Re: Thermodynamics B/C

Posted: April 15th, 2018, 5:06 pm
by UTF-8 U+6211 U+662F
Justin72835 wrote:
UTF-8 U+6211 U+662F wrote: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]
Your turn!

Re: Thermodynamics B/C

Posted: April 25th, 2018, 2:35 pm
by MattChina
i guess ill restart this.
What is the point at which solid, liquid and gas phases are in equilibrium called?