Change in temperature of air in a closed system

In summary, the final temperature of the air inside the glass is impossible based on the information given.
  • #1
Homework Statement
Assume you have a candle inside a drinking glass (cylinder turned upside down) that burns for
10 s.
You know the following values.

Air density = 1.2 kg/m^3
Diameter of the cylinder = 6,5 cm
Height of the cylinder = 6 cm
Initial temperature = 293,15 K
Heating power of the candle = 30 W
Time = 10 s
Specific heat capacity of air = 1 KJ/kg*K


Calculate the final temperature of the air inside the glass.
Relevant Equations
c = Q/m*ΔT
Q = P*t
Looking at the given values, I thought the specific heat formula could be used to calculate the final temperature of the air: c = Q/mΔT. Since the final temperature is the sum of the initial temperature and the change in temperature, the formula can be rearranged to ΔT = Q/m*c.
Q = 30 W * 10 s = 300 J
V = A * h = (π * r^2) * 6 cm = (π * 3,25 cm^2) * 6 cm = 33,2 cm^2 * 6 cm = 199 cm^3
m = V * p = 0,000199 m^3 * 1,2 kg/m^3 = 0,00024 kg
c = 1 KJ/kg*K

ΔT = Q/m*c = 0,3 KJ/(0,00024 kg * 1KJ/kg*K) = 1250 K
Final temperature = 1250 K + 293,15 K = 1543,15 K
This number seems physically impossible, so I conducted the experiment with an initial temperature of 293,15 K, the same air volume and a normal tealight that burned for approximately 10 s. The oven thermometer inside showed a final temperature of about 304,15 K, meaning ΔT = 11K. What did I do wrong?
 
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  • #2
Thomas_Kellner said:
Homework Statement:: Assume you have a candle inside a drinking glass (cylinder turned upside down) that burns for
10 s.
You know the following values.

Air density = 1.2 kg/m^3
Diameter of the cylinder = 6,5 cm
Height of the cylinder = 6 cm
Initial temperature = 293,15 K
Heating power of the candle = 30 W
Time = 10 s
Specific heat capacity of air = 1 KJ/kg*KCalculate the final temperature of the air inside the glass.
Relevant Equations:: c = Q/m*ΔT
Q = P*t

Looking at the given values, I thought the specific heat formula could be used to calculate the final temperature of the air: c = Q/mΔT. Since the final temperature is the sum of the initial temperature and the change in temperature, the formula can be rearranged to ΔT = Q/m*c.
Q = 30 W * 10 s = 300 J
V = A * h = (π * r^2) * 6 cm = (π * 3,25 cm^2) * 6 cm = 33,2 cm^2 * 6 cm = 199 cm^3
m = V * p = 0,000199 m^3 * 1,2 kg/m^3 = 0,00024 kg
c = 1 KJ/kg*K

ΔT = Q/m*c = 0,3 KJ/(0,00024 kg * 1KJ/kg*K) = 1250 K
Final temperature = 1250 K + 293,15 K = 1543,15 K
This number seems physically impossible, so I conducted the experiment with an initial temperature of 293,15 K, the same air volume and a normal tealight that burned for approximately 10 s. The oven thermometer inside showed a final temperature of about 304,15 K, meaning ΔT = 11K. What did I do wrong?
Isn't most of the heat capacity in the glass and tea light?
 
  • #3
haruspex said:
Isn't most of the heat capacity in the glass and tea light?
So basically, only about 0,00027 KJ of warmth is transferred to the air in the system and the rest is stored in the paraffin wax and glass? How is that possible if the specific heat capacity for paraffin wax is about 2,1 KJ/kg*K and for glass is about 0,7 KJ/kg*K? Also, the warmth given off by the flame would also first have to pass through the air before warming up the glass, right?
 
  • #4
1) The teacups output is likely decreasing as oxygen inside the cup is consumed, how long until it extinguishes itself in that cup?

2) It's a thin clear glass cup? Heat is escaping to the external environment. It doesn't have to go through the air via convection, as most of the heat from the flame is propagating via radiation.

3) Parafin wax was has a pretty low thermal conductivity, so I wouldn't think much heat is stored in there, given the short timescale of the experiment.
 
  • #5
Thomas_Kellner said:
So basically, only about 0,00027 KJ of warmth is transferred to the air in the system and the rest is stored in the paraffin wax and glass? How is that possible if the specific heat capacity for paraffin wax is about 2,1 KJ/kg*K and for glass is about 0,7 KJ/kg*K? Also, the warmth given off by the flame would also first have to pass through the air before warming up the glass, right?
What is the mass of the glass? How much will that warm from 300J?
 
  • #6
erobz said:
1) The teacups output is likely decreasing as oxygen inside the cup is consumed, how long until it extinguishes itself in that cup?

2) It's a thin clear glass cup? Heat is escaping to the external environment. It doesn't have to go through the air via convection, as most of the heat from the flame is propagating via radiation.

3) Parafin wax was has a pretty low thermal conductivity, so I wouldn't think much heat is stored in there, given the short timescale of the experiment.
1) The flame extinguishes itself after 10 seconds.
2) It’s relatively thick which is why we assume the system is closed. You mentioned radiation, does that mean a more appropriate way of calculating the final temperature is via the Stefan Boltzmann law? Also, what role does convection play?
3) Yes, your right, but glass has a higher thermal conductivity than air.
 
  • #7
haruspex said:
What is the mass of the glass? How much will that warm from 300J?
The mass is about 400 g. So the increase of temperature would be : 0,3 KJ/0,4 kg*0,72 KJ/kg*K = 1,04 K
 
  • #8
Something I read ( I don't know if I should post it - its not peer reviewed) say that glass is very transparent to wavelengths under ##2.7 \rm{\mu m}##, and candle flame (appearing mostly yellow to the eye) is under ##0.7 \rm{\mu m}##.

Based on that I wouldn't expect the glass to have significantly warmed. Do you have a thermocouple on the glass exterior to see if it's warming significantly?
 
  • #9
Thomas_Kellner said:
1) The flame extinguishes itself after 10 seconds.
2) It’s relatively thick which is why we assume the system is closed. You mentioned radiation, does that mean a more appropriate way of calculating the final temperature is via the Stefan Boltzmann law? Also, what role does convection play?
3) Yes, your right, but glass has a higher thermal conductivity than air.
1) This indicates that the power output is not constant ##30 \rm W##.

2) Radiation is very likely escaping directly to the external environment without warming the glass.

The flame will warm the surrounding air by convection too. The glass having a higher thermal conductivity will conduct heat across itself to the external environment via convection.

The glass will have some amount of heat stored in its mass, how much is a difficult question.

All these "real considerations" should lessen the measured temperature of the air inside.
 
  • #10
Thomas_Kellner said:
The oven thermometer inside ##\dots##
Oven thermometer? I dragged out my oven thermometer and weighed it on my kitchen scale to get 46 g. Assuming that all of it is stainless steel, its specific heat is 0.5 J/(g⋅°C). To raise its temperature by 11 °C (which is what it recorded), it would absorb ##\Delta Q= 0.5~[\text{J/(g⋅°C)}]*46~[\text{g}]*11~[\text{°C}]=230~[\text{J}]## of whatever heat your candle put out. I am assuming, of course, that your oven thermometer is the same as mine.

The short of it is that it is very unrealistic to assume that all the heat that the candle puts out goes into raising the temperature of the air inside. Maybe that is the point of this problem.
 
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  • #11
erobz said:
Something I read ( I don't know if I should post it - its not peer reviewed) say that glass is very transparent to wavelengths under ##2.7 \rm{\mu m}##, and candle flame (appearing mostly yellow to the eye) is under ##0.7 \rm{\mu m}##.

Based on that I wouldn't expect the glass to have significantly warmed. Do you have a thermocouple on the glass exterior to see if it's warming significantly?
It’s not warming significantly. But how is that going to help calculating the final temperature of the air inside the glass?
 
  • #12
Thomas_Kellner said:
It’s not warming significantly.
If it's not warming significantly, then it's not storing much energy.
Thomas_Kellner said:
But how is that going to help calculating the final temperature of the air inside the glass?
Its not really going to help. In theory, if you knew the convection coefficients ( internal\external), surface temperature and glass geometry you could determine the air temperature inside the glass based off the heat flux through the glass walls. To determine the convection coefficients is a complex problem in and of itself.

Like I said, it does help to narrow down what heat is stored where though. So will confirming what @kuruman has suggested about the oven thermometer being a heat sink. Are you using a thermometer type that is absorbing significant heat to capture the measurement?
 
  • #13
Thomas_Kellner said:
The mass is about 400 g. So the increase of temperature would be : 0,3 KJ/0,4 kg*0,72 KJ/kg*K = 1,04 K
I didn't catch this... glass has a pretty high specific heat. So maybe what is a significant change in temperature of the glass needs more careful examination too?
 
  • #14
As far as the mode in which the air in the glass is warmed, I suspect that's predominantly achieved by the active combustion process as opposed to passive convective heat flow or radiation heat transfer (I believe clear air is pretty transparent to those wavelengths too)?
 
  • #15
Thomas_Kellner said:
It’s not warming significantly.
You calculated that, even if all 300J went into warming the glass, it would only go up 1K. On top of that, we have some fraction of the energy going straight through the glass as radiation (including some IR), the thermometer as a heat sink, and the glass losing a little heat to the surrounding air.
If the glass warms by 0.25K it would be significant in the conext of the calculation you are trying to do.
 
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  • #16
haruspex said:
You calculated that, even if all 300J went into warming the glass, it would only go up 1K. On top of that, we have some fraction of the energy going straight through the glass as radiation (including some IR), the thermometer as a heat sink, and the glass losing a little heat to the surrounding air.
If the glass warms by 0.25K it would be significant in the conext of the calculation you are trying to do.
In addition the output power of the candle is declining as air/fuel is converted to combustion products in the closed system.

Overall, it seems like this experiment is far from controlled.
 
  • #17
erobz said:
air/fuel is converted to combustion products in the closed system.
Likely raising the specific heat capacity of the 'air'.
 

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