## Thermodynamics B/C

Justin72835
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### Re: Thermodynamics B/C

Hmm I got 187 J/K as my answer when I did it. I think the fourth line should be

$T(Q) = 363.15 K - \frac{Q}{14360 J/K}$

with a minus instead of a plus. If it were plus, then you would see your temperature increasing (since Q is always positive) instead of decreasing. I redid your work with this correction and I got the same answer. What do you think?
$\frac{Q}{\Delta T} = 3.5 L * 1 kg/L * 4180 J/(kg*K) = 14630 J/K$

$Q_f = 14630 J/K * 25 K = 365750 J$

$\Delta T(Q) = -\frac{Q}{14360 J/K}$

$T(Q) = 363.15 K - \frac{Q}{14360 J/K}$

$\Delta S_{total} = \Delta S_{surroundings} + \Delta S_{water} = \frac{365750 J}{297.15 K} - \int^{365750 J}_{0 J} \frac{dQ}{T(Q)}$

$\Delta S_{total} = 1230.856 J/K - 1044.22 J/K = 186.634 J/K \approx 190 J/K$
EDIT: Oh yeah, next question. An 10g iron rod at 50 degrees Celsius is dropped into a 1L beaker of water at 60 degrees Celsius. How does the length of the rod change? Use the values of c = 0.50 J/(g*K) for iron and c = 4.0 J/(g*K) for water. Use as many sigfigs as you want.
First find the equilibrium temperature:

$10*0.5*(T_f-50)=-1000*4.0*(T_f-60)$ which gives an equilibrium temperature of 59.9875 degrees Celsius

The change in temperature if Tf - Ti = 9.98752 degrees Celcius. Now we can put this into the linear expansion equation:

$\frac{\Delta L}{L}=\alpha \Delta T=12*10^{-6} *9.98752=0.00012$

So the length of the iron rod changes by 0.012%
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### Re: Thermodynamics B/C

Correct (as far as I know). Your turn!

Justin72835
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### Re: Thermodynamics B/C

Correct (as far as I know). Your turn!
Alright this question has two parts. The chemical compound ethanol has an enthalpy of combustion of -1360 kJ/mol. You have a piston that can expand and contract. Assume that the lid of the piston is massless and the gas has a pressure of 1 atm. There are 5.2 moles of ideal gas in the piston at 730 K, and you apply enough heat such that the gas expands isothermally until its pressure is only 60% of the original pressure.

1. How much work was done by the gas?

2. Assuming that the heat was added to the gas during this process was formed from the combustion of ethanol, find the number of moles of ethanol used in the reaction.
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But in ourselves, that we are underlings."

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### Re: Thermodynamics B/C

Alright this question has two parts. The chemical compound ethanol has an enthalpy of combustion of -1360 kJ/mol. You have a piston that can expand and contract. Assume that the lid of the piston is massless and the gas has a pressure of 1 atm. There are 5.2 moles of ideal gas in the piston at 730 K, and you apply enough heat such that the gas expands isothermally until its pressure is only 60% of the original pressure.

1. How much work was done by the gas?

2. Assuming that the heat was added to the gas during this process was formed from the combustion of ethanol, find the number of moles of ethanol used in the reaction.
1. None if there is no atmospheric pressure or other resisting force
2. PV = k; If P multiples by .6, then V divides by .6, or multiplies by 5/3. The work done on the gas is PdV, or nRT*ln(V2/V1) = 5.2 mol * 8.314 J/(mol*K) * 730 K * ln(5/3) = 16122 J = 16.122 kJ. 16.122 kJ / (1360 kJ/mol) = [b].0119 moles[/b].
P.S. Is it just me or should we have an
 tag?

Justin72835
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### Re: Thermodynamics B/C

1. None if there is no atmospheric pressure or other resisting force
2. PV = k; If P multiples by .6, then V divides by .6, or multiplies by 5/3. The work done on the gas is PdV, or nRT*ln(V2/V1) = 5.2 mol * 8.314 J/(mol*K) * 730 K * ln(5/3) = 16122 J = 16.122 kJ. 16.122 kJ / (1360 kJ/mol) = [b].0119 moles[/b].
P.S. Is it just me or should we have an
 tag?
Yup that answer is correct! Also, I agree that there should be a specific tag for question marathons. Your turn!
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### Re: Thermodynamics B/C

What is the heat capacity of 50 kg of $H_2O$ at 0 K? What is the heat capacity of 50 kg of $H_2O$ at 273.15 K?

Justin72835
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### Re: Thermodynamics B/C

What is the heat capacity of 50 kg of $H_2O$ at 0 K? What is the heat capacity of 50 kg of $H_2O$ at 273.15 K?
I'm pretty stumped with this question so I'm just going to make an educated guess.

With the first scenario, the H2O is in the form of ice and has a specific heat of 2.1 J/gK. Multiplying this value by 50,000 (since there are 1000 g per kg) gives a heat capacity of 105 kJ/K. For the second scenario, I'm going to assume that the H2O is in its liquid form and has a specific heat of 4.8 J/gK. Doing the same procedure gives a heat capacity of 240 kJ/K.

If these answers turn out to be completely wrong then I'm going to guess that there is no solution for either of the scenarios, because water can't reach 0 K and at 273.15 K water does not experience a change in temperature when heated or cooled.
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### Re: Thermodynamics B/C

What is the heat capacity of 50 kg of $H_2O$ at 0 K? What is the heat capacity of 50 kg of $H_2O$ at 273.15 K?
I'm pretty stumped with this question so I'm just going to make an educated guess.

With the first scenario, the H2O is in the form of ice and has a specific heat of 2.1 J/gK. Multiplying this value by 50,000 (since there are 1000 g per kg) gives a heat capacity of 105 kJ/K. For the second scenario, I'm going to assume that the H2O is in its liquid form and has a specific heat of 4.8 J/gK. Doing the same procedure gives a heat capacity of 240 kJ/K.

If these answers turn out to be completely wrong then I'm going to guess that there is no solution for either of the scenarios, because water can't reach 0 K and at 273.15 K water does not experience a change in temperature when heated or cooled.
The answer to 1 is actually 0, and 2 was a trick question, as it had [i]two[/i] answers, one for liquid water and one for ice.
You can go though

Justin72835
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### Re: Thermodynamics B/C

What is the heat capacity of 50 kg of $H_2O$ at 0 K? What is the heat capacity of 50 kg of $H_2O$ at 273.15 K?
I'm pretty stumped with this question so I'm just going to make an educated guess.

With the first scenario, the H2O is in the form of ice and has a specific heat of 2.1 J/gK. Multiplying this value by 50,000 (since there are 1000 g per kg) gives a heat capacity of 105 kJ/K. For the second scenario, I'm going to assume that the H2O is in its liquid form and has a specific heat of 4.8 J/gK. Doing the same procedure gives a heat capacity of 240 kJ/K.

If these answers turn out to be completely wrong then I'm going to guess that there is no solution for either of the scenarios, because water can't reach 0 K and at 273.15 K water does not experience a change in temperature when heated or cooled.
The answer to 1 is actually 0, and 2 was a trick question, as it had [i]two[/i] answers, one for liquid water and one for ice.
You can go though
I guess I still have a bit of studying to do

You have a room with dimensions 4 meter by 13 meters by 9 meters. The room is filled with 46 kg of an ideal gas which is at a temperature of 22 °C. After pumping out some air, you find that the pressure of the room dropped by half and the temperature of the room dropped by 13 °C.

Find the mass of the air pumped out of the room.
"The fault, dear Brutus, is not in our stars,
But in ourselves, that we are underlings."

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Riptide
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### Re: Thermodynamics B/C

I'm pretty stumped with this question so I'm just going to make an educated guess.

With the first scenario, the H2O is in the form of ice and has a specific heat of 2.1 J/gK. Multiplying this value by 50,000 (since there are 1000 g per kg) gives a heat capacity of 105 kJ/K. For the second scenario, I'm going to assume that the H2O is in its liquid form and has a specific heat of 4.8 J/gK. Doing the same procedure gives a heat capacity of 240 kJ/K.

If these answers turn out to be completely wrong then I'm going to guess that there is no solution for either of the scenarios, because water can't reach 0 K and at 273.15 K water does not experience a change in temperature when heated or cooled.
The answer to 1 is actually 0, and 2 was a trick question, as it had [i]two[/i] answers, one for liquid water and one for ice.
You can go though
I guess I still have a bit of studying to do

You have a room with dimensions 4 meter by 13 meters by 9 meters. The room is filled with 46 kg of an ideal gas which is at a temperature of 22 °C. After pumping out some air, you find that the pressure of the room dropped by half and the temperature of the room dropped by 13 °C.

Find the mass of the air pumped out of the room.
Sorry to interrupt your guys's chain of questions but I can finally answer one! Since we are dealing with an ideal gas, we can use the ideal gas law which states PV = nRT. Volume and R stay constant, so we can ignore those. The initial ratio of P/nT must equal the final ratio of P/nT. Lets assume we start with a pressure of 1 to keep the math simple. This gives us an equation 1/(46*(273+22)) = .5/(n*(273+22-13)). Solving for n gives us a final mass of 24 kg for the air. The question asks for how much is removed, so subtracting 24.1 from 46 gives us 22 kg of air pumped out of the room.
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