Changing the Air Temperature with a hair dryer

  • #1
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Homework Statement:
A hair dryer blows air at 10 Liters per second, the air is heated with a 1000 W heater while the air is measured at 960 mbar and 20 Degrees Celsius. What is the Temperature of the air when it is leaving the hair dryer.
Relevant Equations:
KE = 3/2kT
Hi, so I found this on another old "AP" High School Finals Exam.
I think I may be super lost.
Because the only way that I can think about is KE = 3/2kT. And then that the difference of the Kinetic Energy of the air Particles is the Q supplied by the heater inside the air dryer.
So ## \frac {3}{2}k*(T_1 - T_2) = 1000 W ## then ##T_2 = T_1 -\frac {2}{3k}*1000 ##

Or with ## Q = mc(T_1 - T_2) ## but we'd need a table for c and the density of air then. Which are then usually supplied.

I think this is wrong because I am also given the Pressure and Volume and I wouldn't need it then. But I have been wrecking my brain about for a bit and I can't figure out where. I'm so sorry.

Thank you for any pointers that could help push me in the right direction.
 

Answers and Replies

  • #2
mjc123
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Use the ideal gas equation to calculate the number of moles in 10 L at the given P and T. What is the molar heat capacity of a diatomic gas?
 
  • #3
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Use the ideal gas equation to calculate the number of moles in 10 L at the given P and T. What is the molar heat capacity of a diatomic gas?
Hi mjc123 Thank you for answering.
My first intuition was actually to use the Ideal Gas Law and then go with the heat capacity.
The number of mols are 0.394... by my calculation. I was able to look up the heat capacity. It's 1.4 Edit! (1.4*3/2*R)?
But I wouldn't have been able to know it by heart, which led me to think I should consider another approach as they usually give you all the constants that you'll need.

Edit: Hmm, I have another table and the molar heat capacity in this book for air is actually 29.1?
 
  • #4
mjc123
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Cv for a diatomic gas (with a high vibrational frequency) is 5/2*R as there are 5 degrees of freedom, 3 translational and 2 rotational. Cp (which is what you need here) is Cv + R = 7/2*R = 29.1 J/mol/K. Cp/Cv = 7/5 = 1.4 for the diatomic gas. These are things you should know without being given them, if you have studied ideal gases.
 
  • #5
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Cv for a diatomic gas (with a high vibrational frequency) is 5/2*R as there are 5 degrees of freedom, 3 translational and 2 rotational. Cp (which is what you need here) is Cv + R = 7/2*R = 29.1 J/mol/K. Cp/Cv = 7/5 = 1.4 for the diatomic gas. These are things you should know without being given them, if you have studied ideal gases.
Hi mjc123.
Thank you, that is very helpful to know. I don't remember in which textbook I got it (I think the molar heat capacity way wasn't in many) from but I'll search and find it again to commit to memory.
 
  • #6
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Cv for a diatomic gas (with a high vibrational frequency) is 5/2*R as there are 5 degrees of freedom, 3 translational and 2 rotational. Cp (which is what you need here) is Cv + R = 7/2*R = 29.1 J/mol/K. Cp/Cv = 7/5 = 1.4 for the diatomic gas. These are things you should know without being given them, if you have studied ideal gases.
Hi mjc123,

So I worked out the problem and I got around 107 Degrees C. Which seems about right (and I get almost the same by looking up the average density of the air and multiplying by Volume to use mc(T_1-T_2).

Just plugging in the numbers my first intuition is definitely wrong and very unreasonable.

I still wonder though. Isn't 3/2kT_1 - 3/2kT_2 the difference in kinetic energy?
 
  • #7
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Hi

I solved it with the Molar Heat Capacity thanks to mjc123's help and got a difference of 87.2 Degrees.

I've been trying yesterday and today morning to verify my results with ##m*c*(T_1-T_2)## and ##3/2*n*R*(T_1-T_2)##

For the former, I used the density of ## 1.293 \frac {kg} {m^3} ## For R I used 8.314 and I've gotten two completely different results again.
Why wouldn't these two work?
 
  • #8
21,483
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Hi

I solved it with the Molar Heat Capacity thanks to mjc123's help and got a difference of 87.2 Degrees.

I've been trying yesterday and today morning to verify my results with ##m*c*(T_1-T_2)## and ##3/2*n*R*(T_1-T_2)##

For the former, I used the density of ## 1.293 \frac {kg} {m^3} ## For R I used 8.314 and I've gotten two completely different results again.
Why wouldn't these two work?
The R you used is in units of 1/moles and the density is in kg
 
  • #9
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The R you used is in units of 1/moles and the density is in kg
Thank you chestermiller for answering.
I'm sorry I don't quite understand?
Do you mean there is a way to equate these two and arrive at a solution?

I didn't use these in the same equation.
Rather in two ways, in which I thought I should arrive at the same solution.

The Temperature Difference.

## (T_1 - T_2) = \frac {Q}{p*V*c} = \frac {2Q}{3*n*R*T} ##

how to write delta T in Latex?
 
  • #10
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The molar density of the air is ##\frac{P}{RT}=\frac{(96000J)(mole)(K)}{m^3(8.314J)(293K)}=39.4moles/m^3=0.0394moles/l##This is the same as (39.4)(29)/1000 = 1.143 kg/m^3, so you already used the wrong density.

The molar flow rate of air is (0.0394)(10)=0.394 moles/second.

The molar heat capacity at constant pressure of air is 3.5 R = 29.1 J/mole. $$\Delta T = \frac{1000}{(0.394)(29.1)}=87.2 C$$

My point is that, when you do this calculation, you have to make sure that the units are consistent.

You keep using 1.5R for the molar heat capacity of the air. This is incorrect.

In Latex, you write delta as backslash Delta.
 
Last edited:
  • #11
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Thank you chestermiller.

There might be a misunderstanding.
I did use the correct formula and R.
And I've gotten the same result as you. I mentioned this above, but I probably wasn't clear enough, I'm sorry.

I tried to verify that result ## \Delta T= 87.2C ## by using ## Q=m*c* \Delta T## and ## Q = 3/2 n*R*\Delta T ## Thanks to your pointer I was able to verify with the second equation, thank you! (if I substitute 7/2 R for 3/2R here I get the correct result).

But the former is still way off if I substitute the density of air 1.293 times the volume in cubic meters which should yield kg's for mass no?
 
  • #12
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Thank you chestermiller.

There might be a misunderstanding.
I did use the correct formula and R.
And I've gotten the same result as you. I mentioned this above, but I probably wasn't clear enough, I'm sorry.

I tried to verify that result ## \Delta T= 87.2C ## by using ## Q=m*c* \Delta T## and ## Q = 3/2 n*R*\Delta T ## Thanks to your pointer I was able to verify with the second equation, thank you! (if I substitute 7/2 R for 3/2R here I get the correct result).

But the former is still way off if I substitute the density of air 1.293 times the volume in cubic meters which should yield kg's for mass no?
Like I said, the density is not 1.293 kg/m^3. It is 1.143 kg/m^3. And, at 10 liters/sec, this represents a mass flow rate of 0.01143 kg/sec. The specific heat of air is 29.1 J/mole -K= 1.0034 J/gm-K = 1003.4 J/kg-K. So, mc = (0.01143)(1003.4)=11.46 J/K. So, $$\Delta T = \frac{1000}{11.46}= 87.2 C$$
 
  • #13
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Like I said, the density is not 1.293 kg/m^3. It is 1.143 kg/m^3. And, at 10 liters/sec, this represents a mass flow rate of 0.01143 kg/sec. The specific heat of air is 29.1 J/mole -K= 1.0034 J/gm-K = 1003.4 J/kg-K. So, mc = (0.01143)(1003.4)=11.46 J/K. So, $$\Delta T = \frac{1000}{11.46}= 87.2 C$$
Thank you chestermiller.
May I ask if you calculated the density and if so how? All I knew how to do was to look it up in a textbook table.
 
  • #15
mjc123
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I still wonder though. Isn't 3/2kT_1 - 3/2kT_2 the difference in kinetic energy?
3/2*kT is the average translational kinetic energy of the molecules. But (apart from monatomic gases) this is not the only form of energy they can have. Specifically, a diatomic molecule can rotate about 2 perpendicular axes, so it has two rotational degrees of freedom, each of which has an average energy of 1/2*kT. Therefore the molar Cv is 5/2*R, and Cp is 7/2*R.
 
  • #16
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Post #10 shows how I used the ideal gas law to determine the air density.
Thank you. I saw what you did in the first part and I could follow. I have to try to reason a bit on how we arrived at the density, I'm don't follow this quite yet.
 
  • #17
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3/2*kT is the average translational kinetic energy of the molecules. But (apart from monatomic gases) this is not the only form of energy they can have. Specifically, a diatomic molecule can rotate about 2 perpendicular axes, so it has two rotational degrees of freedom, each of which has an average energy of 1/2*kT. Therefore the molar Cv is 5/2*R, and Cp is 7/2*R.
Thank you, this sounds very intriguing.
I wasn't able to find any textbook that explained this way of calculating C_p yet. But I'm going to search further.
Thanks again.
 
  • #18
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Thank you. I saw what you did in the first part and I could follow. I have to try to reason a bit on how we arrived at the density, I'm don't follow this quite yet.
$$\rho=\frac{m}{V}=\frac{nM}{V}=M\frac{n}{V}$$
$$\frac{n}{V}=\frac{p}{RT}$$
 
  • #19
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$$\rho=\frac{m}{V}=\frac{nM}{V}=M\frac{n}{V}$$
$$\frac{n}{V}=\frac{p}{RT}$$
Thank you! It took a while to wrap my head around it but I can completely follow now what you did. Thank you for your patience. I was especially slow today (hopefully it's just the heat!) and yet I learned a new relation that I think will come in handy!
 
  • #20
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Cv for a diatomic gas (with a high vibrational frequency) is 5/2*R as there are 5 degrees of freedom, 3 translational and 2 rotational. Cp (which is what you need here) is Cv + R = 7/2*R = 29.1 J/mol/K. Cp/Cv = 7/5 = 1.4 for the diatomic gas. These are things you should know without being given them, if you have studied ideal gases.
Hi mjc123, sorry to come back to this thread again after such a long time.
thanks to you and chestermiller I think I have a real good grasp and can use these formulas well (I did a lot of problems I found in two textbooks).

I still have a concept question though and it might be a silly question, for which if it is i am sorry.
Why (If I understood correctly) does the temperature only depend on translational kinetic energy and not rotational kinetic energy?
 
  • #21
DrClaude
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Why (If I understood correctly) does the temperature only depend on translational kinetic energy and not rotational kinetic energy?
Temperature is related to all the ways in which a system can store energy, not just kinetic energy.
 
  • #22
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Temperature is related to all the ways in which a system can store energy, not just kinetic energy.
Hi, thanks drclaude.
I'm sorry if I'm misunderstanding something here.
I was just wondering since the kinetic energy is ## \frac {1}{2}m(v_rms)^2 = \frac {3}{2}kT ## it seemed to me like temperature is only dependent on kinetic energy.
And I was wondering why there isn't accounted for rotational energy like with the molar heat capacity. when we account for rotational energy as well in diatomic or polyatomic molecules?
 
  • #23
mjc123
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Why (If I understood correctly) does the temperature only depend on translational kinetic energy and not rotational kinetic energy?
It might be more helpful if you think about it the other way round, that energy depends on temperature (though it comes to the same thing in the end). Have you come across the principle of equipartition of energy? It says that the energy is (on average) equally distributed between all the available degrees of freedom. [Actually, this is true only when the separation of energy levels << kT, so the energy variation is effectively continuous. That's why, for diatomic gases at normal temperatures, we include the rotations but not the vibration, for which ΔE >> kT.]

Thus for a mole of monatomic molecules, if you add at constant volume an amount of heat equal to 15R, this will be equally divided between the 3 translational degrees of freedom, with 5R in each, so the temperature rises by 10K (the heat capacity being R/2 per degree of freedom). For a diatomic gas, the energy will be divided between the 5 degrees of freedom, each getting 3R, so the temperature will rise by 6K.
 
  • #24
DrClaude
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I'm sorry if I'm misunderstanding something here.
I was just wondering since the kinetic energy is ## \frac {1}{2}m(v_rms)^2 = \frac {3}{2}kT ## it seemed to me like temperature is only dependent on kinetic energy.
And I was wondering why there isn't accounted for rotational energy like with the molar heat capacity. when we account for rotational energy as well in diatomic or polyatomic molecules?
You have to see the relation the other way around. At a given temperature, you can use ##\frac {3}{2}kT ## to figure out the average translation kinetic energy. You could do the same for the rotational energy, electronic energy, etc.

Temperature is defined for a system at equilibrium. In that case, the equipartition theorem can be used to figure out how its thermal energy is partitioned among the different degrees of freedom.

Note that there is an unstated assumption in the discussion in this thread, namely that the temperature is close to STP. At higher temperature, molecular vibrations will also become excited and will enter into the calculation for heat capacity.
 
  • #25
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Hi mjc123 and drclaude.
Thank you for explaining. I have come across the equipartition theorem in halliday's textbook (which I got after this problem because I haven't come across the various C_v before).

I think I have a better idea now, but I'll have to ponder on it a bit more to really understand.

Thanks again
 

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