## Thermodynamics B/C

Justin72835
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### Re: Thermodynamics B/C

UTF-8 U+6211 U+662F wrote:
Justin72835 wrote:
I'm pretty sure that for this problem you would just use an equation to convert between the two temperatures.

For converting from Kelvin to Réaumur: R = (K - 273.15)*0.8

For converting from Réaumur to Kelvin: K = 1.25R + 273.15
You have a heat refrigerator that requires 1340 W of work. You also find that the refrigerator releases heat into a heat sink at a rate of 4250 W. How much heat can the refrigerator extract from a cold reservoir and what is its coefficient of performance?
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### Re: Thermodynamics B/C

Justin72835 wrote: You have a heat refrigerator that requires 1340 W of work. You also find that the refrigerator releases heat into a heat sink at a rate of 4250 W. How much heat can the refrigerator extract from a cold reservoir and what is its coefficient of performance?
First Law of Thermodynamics: $1340 J + Q_c = 4250 J$. $Q_c = 2910 J$. $COP = \frac{Q_h}{W} = \frac{4250 J}{1340 J} = 3.17$. [b]2910 W[/b] and [b]3.17[/b]

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### Re: Thermodynamics B/C

UTF-8 U+6211 U+662F wrote:
Justin72835 wrote: You have a heat refrigerator that requires 1340 W of work. You also find that the refrigerator releases heat into a heat sink at a rate of 4250 W. How much heat can the refrigerator extract from a cold reservoir and what is its coefficient of performance?
First Law of Thermodynamics: $1340 J + Q_c = 4250 J$. $Q_c = 2910 J$. $COP = \frac{Q_h}{W} = \frac{4250 J}{1340 J} = 3.17$. [b]2910 W[/b] and [b]3.17[/b]
Overall correct but since we're dealing with a heat refrigerator the formula is actually COP = Qc/W. This is because we are interested in the heat we can extract (Qc) rather than the heat given off at the very end (Qc). Everything is else was good. Your turn!
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### Re: Thermodynamics B/C

A 500. g iron rod at 50. degrees Celsius is dropped into a (very big) beaker filled with 2.0 L of water at 20. degrees Celsius. 80.0% of the heat transferred from the iron rod is lost to the surroundings and is not transferred to the water. Find the power in watts of the transfer of energy into the water if the system reaches thermal equilibrium in exactly one minute.

Use the values of c = 0.50 J/(g*K) for iron and c = 4.0 J/(g*K) for water. Use appropriate significant figures.

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### Re: Thermodynamics B/C

UTF-8 U+6211 U+662F wrote:A 500. g iron rod at 50. degrees Celsius is dropped into a (very big) beaker filled with 2.0 L of water at 20. degrees Celsius. 80.0% of the heat transferred from the iron rod is lost to the surroundings and is not transferred to the water. Find the power in watts of the transfer of energy into the water if the system reaches thermal equilibrium in exactly one minute.

Use the values of c = 0.50 J/(g*K) for iron and c = 4.0 J/(g*K) for water. Use appropriate significant figures.
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:

$Q_{iron}=-(Q_{water}+Q_{surroundings})=-5Q_{water}$

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

$m_{iron}c_{iron}(T_f-50)=-5m_{water}c_{water}(T_f-20)$ 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.
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### Re: Thermodynamics B/C

Justin72835 wrote:
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:

$Q_{iron}=-(Q_{water}+Q_{surroundings})=-5Q_{water}$

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

$m_{iron}c_{iron}(T_f-50)=-5m_{water}c_{water}(T_f-20)$ 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!

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### Re: Thermodynamics B/C

UTF-8 U+6211 U+662F wrote: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).
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### Re: Thermodynamics B/C

Justin72835 wrote:
UTF-8 U+6211 U+662F wrote: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).
$\frac{Q}{\Delta T} = 3.5 L * 1 kg/L * 4180 J/(kg*K) = 14630 J/K$

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

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

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

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

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

With special thanks to WolframAlpha.
P.S. is there a better way to format the equations than having a separate $$$ tag every line? Justin72835 Member Posts: 175 Joined: June 25th, 2017, 7:06 am Division: Grad State: TX Has thanked: 0 Been thanked: 0 ### Re: Thermodynamics B/C UTF-8 U+6211 U+662F wrote: [math]\frac{Q}{\Delta T} = 3.5 L * 1 kg/L * 4180 J/(kg*K) = 14630 J/K$

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

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

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

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

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

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

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

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?

Also, we'll have to make do with putting the math tag before every equation
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### Re: Thermodynamics B/C

Justin72835 wrote: 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.

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