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Re: Thermodynamics B/C

Posted: February 11th, 2018, 9:12 am
by UTF-8 U+6211 U+662F
Since the 80% heat lost by the iron rod goes into the surroundings and 20% of it goes into changing the water's temperature, we get the following equation:

[math]Q_{iron}=-(Q_{water}+Q_{surroundings})=-5Q_{water}[/math]

Now we can plug in all our known values to find the equilibrium temperature, then the heat transferred:

[math]m_{iron}c_{iron}(T_f-50)=-5m_{water}c_{water}(T_f-20)[/math] where Tf = 20.1863 degrees Celsius.

Plugging this value back into the equation, you find that the Qiron = 7453.42 J. Dividing this value by 60 gives P = 124 W.
Yes, except transfer of heat into the water and not out of the rod and also only two sigfigs. Your turn!

Re: Thermodynamics B/C

Posted: February 11th, 2018, 12:11 pm
by Justin72835
Yes, except transfer of heat into the water and not out of the rod and also only two sigfigs. Your turn!
You fill a container with 3.5 L of water at an initial temperature of 90 °C. After 1 hour of cooling, you measure the temperature the temperature of the water to have dropped to 65 °C. Lastly, the surrounding temperature of the container remained at a steady 24 °C throughout the entire cooling process. Calculate the change in entropy for the entire system (including both the water and the surroundings).

Re: Thermodynamics B/C

Posted: February 11th, 2018, 1:54 pm
by UTF-8 U+6211 U+662F
Yes, except transfer of heat into the water and not out of the rod and also only two sigfigs. Your turn!
You fill a container with 3.5 L of water at an initial temperature of 90 °C. After 1 hour of cooling, you measure the temperature the temperature of the water to have dropped to 65 °C. Lastly, the surrounding temperature of the container remained at a steady 24 °C throughout the entire cooling process. Calculate the change in entropy for the entire system (including both the water and the surroundings).
[math]\frac{Q}{\Delta T} = 3.5 L * 1 kg/L * 4180 J/(kg*K) = 14630 J/K[/math]

[math]Q = 14630 J/K * 25 K = 365750 J[/math]

[math]\Delta T(Q) = \frac{Q}{14360 J/K}[/math]

[math]T(Q) = 363.15 K + \frac{Q}{14360 J/K}[/math]

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

[math]\Delta S_{total} = 1230.856 J/K - 973.409 J/K = 257.45 J/K \approx 260 J/K[/math]

With special thanks to WolframAlpha.
P.S. is there a better way to format the equations than having a separate tag every line?

Re: Thermodynamics B/C

Posted: February 11th, 2018, 3:16 pm
by Justin72835
[math]\frac{Q}{\Delta T} = 3.5 L * 1 kg/L * 4180 J/(kg*K) = 14630 J/K[/math]

[math]Q = 14630 J/K * 25 K = 365750 J[/math]

[math]\Delta T(Q) = \frac{Q}{14360 J/K}[/math]

[math]T(Q) = 363.15 K + \frac{Q}{14360 J/K}[/math]

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

[math]\Delta S_{total} = 1230.856 J/K - 973.409 J/K = 257.45 J/K \approx 260 J/K[/math]

With special thanks to WolframAlpha.
P.S. is there a better way to format the equations than having a separate tag ---- every line?
Hmm I got 187 J/K as my answer when I did it. I think the fourth line should be



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?

Also, we'll have to make do with putting the math tag before every equation :cry:

Re: Thermodynamics B/C

Posted: February 11th, 2018, 3:31 pm
by UTF-8 U+6211 U+662F
Hmm I got 187 J/K as my answer when I did it. I think the fourth line should be



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?
Yeah, that's my bad.
[math]\frac{Q}{\Delta T} = 3.5 L * 1 kg/L * 4180 J/(kg*K) = 14630 J/K[/math]

[math]Q_f = 14630 J/K * 25 K = 365750 J[/math]

[math]\Delta T(Q) = -\frac{Q}{14360 J/K}[/math]

[math]T(Q) = 363.15 K - \frac{Q}{14360 J/K}[/math]

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

[math]\Delta S_{total} = 1230.856 J/K - 1044.22 J/K = 186.634 J/K \approx 190 J/K[/math]
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.

Re: Thermodynamics B/C

Posted: February 11th, 2018, 5:35 pm
by Justin72835
Hmm I got 187 J/K as my answer when I did it. I think the fourth line should be



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?
Yeah, that's my bad.
[math]\frac{Q}{\Delta T} = 3.5 L * 1 kg/L * 4180 J/(kg*K) = 14630 J/K[/math]

[math]Q_f = 14630 J/K * 25 K = 365750 J[/math]

[math]\Delta T(Q) = -\frac{Q}{14360 J/K}[/math]

[math]T(Q) = 363.15 K - \frac{Q}{14360 J/K}[/math]

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

[math]\Delta S_{total} = 1230.856 J/K - 1044.22 J/K = 186.634 J/K \approx 190 J/K[/math]
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:

[math]10*0.5*(T_f-50)=-1000*4.0*(T_f-60)[/math] 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:

[math]\frac{\Delta L}{L}=\alpha \Delta T=12*10^{-6} *9.98752=0.00012[/math]

So the length of the iron rod changes by 0.012%

Re: Thermodynamics B/C

Posted: February 11th, 2018, 5:37 pm
by UTF-8 U+6211 U+662F
Correct (as far as I know). Your turn!

Re: Thermodynamics B/C

Posted: February 11th, 2018, 5:40 pm
by Justin72835
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.

Re: Thermodynamics B/C

Posted: February 11th, 2018, 5:50 pm
by UTF-8 U+6211 U+662F
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?

Re: Thermodynamics B/C

Posted: February 11th, 2018, 6:08 pm
by Justin72835
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!