Ii.2.2. Stefan'S Constant of Radiation

 

II.2.2. STEFAN’S CONSTANT OF RADIATION 

1. INTRODUCTION:

A hot body emits radiation that will be absorbed by another body. The amount of energy radiated depends on its surface area, the absolute temperature of the body and a property depending on the nature of the surface of the body. This property is called emissivity and is denoted by . This is the ratio of the energy radiated by unit area of the given given surface to that radiated by unit area of a perfectly black surface when both the surfaces are at the same temperature. The energy radiated per second, Q, is given by 4

Q =   A T   (II.2.2.1) Here A is the surface area of the body.  is called the Stefan-Boltzmann constant and is one of the important constants of physics. 

4

3 2

 = ( (  k  / ħ c   (II.2.2.2) k, ħ (= h/2 h/2) and c are the Boltzman constant, the Planck’s constant  constant   divided by 2  2    and the velocity of light in vacuum respectively. A bodythe canStefan’s lose heat to its we surroundings by the conduction, convection radiation. and To measure constant must reduce fraction of heat lost and by conduction convection to a few percent of the total heat lost, so that radiation dominates. This is achieved  by the following steps: (a) blackening the surface of the body and that of the enclosure to make  have the maximum value of nearly unity, (b) increasing the resistance to heat conduction by using very thin wires for the electrical measurements and by using a nylon thread of poor conductivity to suspend the sample and (c) using a closed enclosure to reduce convection losses due to currents of air.

2. EXPERIMENTAL ARRANGEMENT:

B

C

D

A ThC

 N E

H

Figure II.2.2.1: A schematic of the experimental ex perimental arrangement

 

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Figure II.2.2.1 shows a schematic of the set up for measuring Stefan’s constant. A is a light weight aluminum cylinder about 3 cm long and 2 cm in outer diameter with wall thickness of 0.5 mm. To heat the sample we require a heater which can generate about 1 W power. A simple way of making the heater is the following. A Teflon tube of outer diameter 10 mm and wall thickness 0.5 mm is cut to a length of 2.6 cm. A series of five 1mm diameter holes are made around the circumference of the tube at the top and bottom. Five 100-Ohm ¼ W resistors are taken and the leads of each resistor are inserted into the top and bottom holes in the Teflon tube so that the five resistors are connected in parallel and lying flat on their length on the surface of the Teflon tube. tube. All the leads will will be inside the the Teflon tube. All incoming leads are are twisted together and all outgoing leads are twisted together. The total resistance of the heater is 20 (ie. 100/5) ohms and it can generate up to 1 W of heat. A pair of copper leads is soldered soldered to the pair of twisted leads. These are insulated by winding winding a thin Teflon tape around them and are brought out from the top of the aluminum aluminum cylinder. Two or three layers of thin Teflon tape are wound on the outer surface surface of the resistors. The Teflon tube tube is placed within the aluminum aluminum cylinder and the inter-space between the wall of the aluminum cylinder and the outer wall of the Teflon tube is filled with aluminum foil to provide a good thermal link. Heater current leads

Outer Al Cylinder

Teflon cylinder 100  ¼ W resistors connected in parallel Aluminum foil

Figure II.2.2.2: Mounting of the heater in the Al cylinder A copper constantan thermocouple is made by taking about 5 cm of insulated constantan wire (about SW 40) and two long insulated copper wires of the same gauge. After removing the insulation of the wires to a length of 3 to 5 mm at the ends, two copper constantan junctions are made by twisting the bare wires. wires. When the solder is just melting on the soldering iron the two  junctions are pushed into the molten solder and pulled out. ou t. A drop of superglue (a very v ery quick and strong rapid-cure adhesive available in small tubes) is applied on the Aluminum cylinder near the middle where it is desired to place one of the thermocouple junctions. In a few minutes the liquid dries over the surface surface to form form a thin electrically electrically insulating insulating layer. A junction of the thermocouple is placed at this point and a thin insulated copper wire is tied around it to hold it tight against against the cylinder. Another drop of superglue is applied on the thermocouple  junction. One should be careful not to apply too much superglue. If the glue forms a thick layer between the junction and the surface of the aluminium cup, there will be an appreciable temperature drop in this layer layer of glue. The thermocouple will measure measure a temperature lower than the actual temperature of the aluminium surface. After a few minutes the superglue sets and forms a tight joint between the thermocouple and the aluminium cylinder. The aluminum cylinder can then be painted black on the outside with eenamel namel paint. The paint must be allowed to dry. The cylinder is is hung by a thin nylon thread from the top lid of a small small stainless steel can E. The inside inside of the can is also painted black. On the the top lid D banana sockets sockets are fixed

 

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for the current leads and a RCA socket for the thermocouple leads. The second junction of the thermocouple is fixed to the lid D of the stainless steel can E. The copper constantan thermocouple has a thermo-electric power   of 40 micro-volts per degree Kelvin near room temperature. temperature. For a 10-degree difference difference in temperature of the two  junctions the thermo-emf will be about 400 microvolts. This is amplified a hundred times by by the DC amplifier described in Section I. The Stefan’s constant box constant box is shown in Figure II.2.2.3.

Figure II.2.2.3 Stefan’s constant Stefan’s constant box

3. Apparatus required:

Constant current source, DC differential amplifier, Stefan’s Stefan’s constant box, and a multimeter reading in DC 200 mV range. 4. PROCEDURE:

The connection diagram diagram for the Stefan’s constant experiment is shown in Figure II.2.2.4.  II.2.2.4.   A constant current source in the high current mode is connected to the heater terminals of the Stefan’s constant box. The two switches on the constant current source are put in the high current mode. The input terminals I1 of a DC differential amplifier are connected to the RCA socket for thermocouple outlet on the Stefan’s constant box. The band switch on the DC differential amplifier is set at I1  and the toggle switch in the position X100. The output terminals of the DC differential amplifier are connected to a digital multimeter (DMM) set to range DC 200 mV.

 

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     M    M    D

Figure II.2.2.4 Connection diagram for Stefan’s constant experiment ex periment  

Adjust the current through the heater at 200 mA. This is the maximum current you should pass through the heater. Do not exceed this current.  A voltage will appear on the DMM which will increase in magnitude with time. Wait for one hour for the steady steady state state to be reached. Note the value of the output DC voltage in milli- volts V+. (It is assumed that this value is positive. Otherwise toggle the reversing switch to get get a positive voltage). The reversing switch on the differential amplifier is thrown to the other position and the reading V-  is noted. The amplified thermo emf corrected for the offset of the amplifier is Vcorr  =  = (V+ - V-)/2 (II.2.2.3)

Then reduce the current in steps of twenty mA till you you reach 140 mA. current wait for 45 minutes before taking a reading of V+ and V-.

For each value of

The total power dissipated in the resistor is 2

Q=I R

 

(II.2.2.4)

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where I is the current through the heater and R is the resistance of the heater. The resistance of the heater is marked on the Stefan’s Stefan’s constant box. constant box. The temperature difference between the cylinder and the surrounding can is 3

T2  T1 = ( Vcorr  in milli-volts/   )x10 = Vcorr in mV/

(II.2.2.5)

Here  is taken as 40 micro-volts per degree Kelvin and  = 100 (amplification of Differential 3 amplifier). The factor 10  ari  arises ses because we are dividing millivolts by microvolts. microvolts. The room temperature T1 is measured with a mercury thermometer. Knowing T1 and T = (T1T2) from measurement, T2 is calculated. All temperatures should be expressed in Kelvin. If all the heat is lost only by radiation then

4

4

Q =  A (T2   T1 )

(II.2.2.6)

Here  is the the emissivity of the black surface. For different types of black paints the emissivity varies between 0.9 and 0.95, the latter being the value for enamel paints. We take 0.95 as the emissivity of the black surface since we have used enamel paint. The area A of the aluminium cup radiating heat is A = 2 2r (r+l  (r+l ) where r is the radius and l   the length of the the cylinder. stainless steel box. 4

(II.2.2.7) The values of r and l  are   are given on the

4

A graph is plotted with (T 1   –  –  T  T2 ) on the X axis and Q on the Y-axis. A straight line is fitted to the points on a computer and the slope of the straight straight line is is found. The Stefan’s constant is calculated from the slope knowing the area A of the aluminum cylinder. A sample set of readings is given below on Table Table II.2.2.1. II.2.2.1. In this table the Column 1 gives the current current through the heater in milli-amperes. milli-amperes. The heater power Q in watts is given given in column 2. Column 3 gives V+ in  m  mV. V. Column 4 gives V- in mV. Column 5 gives Vcorr  in mV. From this the temperature difference T = T2  –  –  T  T1 is calculated by dividing V corr   by  by    and this is given in column 6. Knowing T1, the temperature T2 of the surface of the aluminum block is 4 4 9 4 calculated and given in column 7. Column 8 gives (T2 T1 ) in units of 10  K  . Figure II.4.2.4 4 4  below gives a plot of Q against (T2 T1 ).

 

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Table II.4.2.1

T1 

299

Dia of Al

K

-2

2x10   m -

Height

3.3x10   m

 Area

2.7x10   m

-

Resistance

amp

20.1

Watt

Ohms

mV

mV

mV

K

K

K 4

4

I

Q

V+  

V- 

Vcorr

T 

T2 

T2 -T1  

0.203

0.8283

180

-176

178

44.5

343.5

5.93E+09

0.182

0.6658

162

-155

158.5

38.66

337.66

5.01E+09

0.161

0.521

127

-123

125

30.49

329.49

3.79E+09

0.141

0.3996

101

-94

97.5

23.78

322.78

2.86E+09

0.121

0.2943

77

-68

72.5

17.68

316.68

2.06E+09

0.101

0.205

54

-48

51

12.44

311.44

1.42E+09

0.9

0.8

-10

4

slope 1.382x10  W/K 0.7

   W   n 0.6    i    Q   r   e   w0.5   o   p   r   e    t 0.4   a   e    H 0.3

0.2 1

2

3

4 4

4

5 9

6

4

(T1 - T2 ) in 10 K

4

4

Figure II.2.2.4 Plot of Q against (T1  –   –  T  T2 )

-10

4

From the figure the slope of the curve is 1.382x10   W/K  . The Stefan’s constant is calculated from the area of surface (given in Table II.2.2.1) and the emissivity of black paint

 

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-8

2

4

taken to to be 0.95. It comes out to be 5.38x10 W/m K   in reasonable agreement with the actual -8 2 4 value 5.67x10  W/m K  . The sources of systematic errors in this experiment are the following: 1.  The emissivity emissivity of the black paint used may be lower than 0.95. important source. 2.  Some heat may be lost by conduction and convection.

This is is the most

If one would like to measure the emissivity of a polished surface, one may have a second set up with a thin copper cylinder with the outer surface electroplated with Nickel. The above above measurements can be repeated. Knowing the value of     the emissivity of the electroplated nickel surface can be found. 5. TIME CONSTANT:

In all heat experiments time constant plays an important role. The temperature grows from its initial value at room temperature to its final value as

T1 + (t) = T1 +    exp  exp ( (t/ t/)]

(II.2.2.8)

where  is (T2-T1), the steady state value of the increase in temperature and   is the relaxation time. Theoretically the steady state temperature is attained after infinite time. But in practice, if one waits for a time t longer than 5 , the temperature is within 1% of its steady state value. It is necessary to understand the factors governing  so that the experiment can be designed with a value of  of the order of 10 minutes. Then the reading reading of the the thermocouple for for a given heater power can be taken after forty five minutes to one hour and the experiment can  be repeated for at least four different heat inputs within three hours. At a time t, part of the heat input goes to raise the temperature of the block and part is lost as radiation, conduction and convection. If the the temperature rise is not too large we may lump all the heat loss mechanisms into a factor C C  This is the Newton’s Newton’s law of cooling.  cooling.   msd msd /dt + C C  Q

(II.2.2.9)

Here ms is the total thermal capacity of the metal block including the thermal capacity of the heater and thermometer. The solution to this equation is  (t) = {Q/C} [1-exp { {  C/ms) C/ms) t}]

(II.2.2.10)

From this equation we identify Q/C as  (  ( ) and (ms/C) as . For reducing the relaxation time  we should reduce the mass of the cup. That is why it is preferable to use a lightweight cup made of aluminum or thin copper.

 

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To calculate what mass we should choose we assume heat is lost only by radiation. radiation. Then 4

4

3

C = A A [(T1+)   T1 ]  4  4 AT AT1  



when  is small. So 3

C  4  4 AT AT1  

(II.2.2.12)

and 3

  ms/(  ms/( 4 4AT AT1 ) -8

2

4

(II.2.2.13) -8

Taking  to be 5.7x10  W/m K  ,  to be 1, T1 to be 300K, A = 2x10  m

2

C  0.0125 So if  is to be less than 600 s, ms < Cx600 = 7.5 J/K The specific heat of aluminum is 0.910 J/gK, means that the mass of the aluminum cylinder should be less than about about 8 gm. A cylinder cylinder of 2 cm diameter, diameter, length length l   of 3 cm and wall 2 thickness t of 0.5 mm will have a volume (2 (2r ll   + + r  )t = 1.09 cc. Density of Aluminum is 2.7 g/cc. The mass of the cylinder in the experiment is about about 2.9 g. To this this must be added the weight of the Teflon cylinder and the five five resistors. This is the reason why you you should wait for at least 45 minutes for steady state to be reached. Questions: 1.  Why is the inside of the wall of a thermos flask silvered? 2.  Liquid nitrogen is stored in in an evacuated double walled stainless stainless steel container. The radius of the the outer of the two walls is 10 cm and its length is 1 meter. The wall is at room temperature (300 K). The container is filled with liquid nitrogen. Its boiling  point is 77 K. Assuming the emissivity of o f stainless steel to be 0.1, calculate the heat reaching the liquid nitrogen from the the outer wall. If the latent heat of vaporization of Liquid nitrogen is 160 KiloJoules/litre of liquid nitrogen calculate the boil off rate of liquid nitrogen. 3. If we put 10 layers of aluminized mylar mylar foils between the outer wall and the inner wall of the doubled wall vessel will it reduce the heat radiation reaching the liquid nitrogen? If the emissivity of the foils is taken to be the same as the emissivity of stainless steel by what factor will the heat reaching liquid nitrogen change?

 

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