> ')&@ D8bjbjFF .:,,$L`n^$JRpQyyyyyb(|r7^00`L &SD
TAP 525-3: Fusion in a kettle?
A change of scale
When you are confident with basic calculations of fission and fusion energy changes, you should work through these questions that try to put the energies of these changes into a more human scale for you. You will also need to understand the conversion of atomic mass units to energy and the meaning of the term electron volt.
Try these
One of the reactions that fuel the stars is the fusion of two protons to give deuterium. In turn the deuterium goes through a series of reactions, the end product being helium. This is also a process that releases energy. In this question you are asked to consider the energy that would be released if all the deuterium in the water contained in an electric kettle were to be converted by fusion into helium.
The kettle contains 1 litre of water. The data you need are listed below.
1 atomic mass unit (u) = 931 MeV
1 eV = 1.6 10 19 J
NA = 6.02 1023 mol 1
ParticleMass / u1.007 8252.014 102EMBED Equation.33.016 030EMBED Equation.31.008 665
1 Two deuterium nuclei EMBED Equation.3can fuse to give one nucleus of helium EMBED Equation.3 with the ejection of one other particle. Write down the balanced equation that represents this reaction.
2 Calculate the mass change that occurs in this reaction.
3 Convert this energy into joules.
This gives you the energy released when two deuterium nuclei fuse. The next steps take you through the calculation of the total energy released if all the deuterium in the kettle water were to fuse to make helium-3. The ratio of deuterium atoms to hydrogen in water is roughly 1 to 7000.
4 What is the mass of 1 mole of water (H = 1 u; O = 16 u roughly)?
5 How many moles of water are contained in the litre?
6 How many molecules of water (H2O) are in the kettle?
7 How many molecules of deuterium oxide (D2O) are in the kettle?
8 Each heavy water molecule has two atoms of deuterium; what total energy is released if all the deuterium in the kettle is converted to helium-3?
Now to put this number in a new perspective. It requires 4200 J to increase the temperature of 1kg of water by 1K.
9 How many litres of water could be heated through 100 K by the fusion energy you calculated in question 8?
Hints
1 It is important to consider the atomic electrons in this equation. You begin with two, one for each hydrogen. How many electrons does an un-ionised atom of deuterium have? So what must one of the emitted particles be? This should lead you to the other particle.
2 The conversions you need are near the data table in the question.
4 The formula of water shows that there are two hydrogen atoms and one oxygen for each water molecule.
5 1 litre of water has a mass of 1 kg.
6 1 mole contains 6 x 1023 molecules of water.
Practical advice
These questions can be modified in many ways, not least by changing the homely example of a kettle to perhaps a bath full of water or even to Lake Windermere or the local reservoir.
Social and human context
The 6000 litres of heated water may not seem so significant until you realise that this has come from the fusion of deuterium which had an original volume of 0.15 cm3.
Answers and worked solutions
1.
EMBED Equation.3
2. Dm = (3.016 030 u + 1.008 665 u) 2 2.014 102 u = 0.0035 u
3. 0.003509 u 931 106 eV u 1 1.6 10 19 J eV 1 = 5.23 10 13 J
4. 18 g
5. 1 litre of water has a mass of 1 kg.
number of04<@hjlnq`G0j-F
h#h.B*CJUV^JaJph!h#h.B*CJ^JaJph*jh#h.B*CJU^JaJph*jh#h.B*CJU^JaJphh#h.5\*jh#h.B*CJU^JaJphh#h.aJh#h.CJEHaJh#h.]h#h.CJEH
aJh#h.OJQJ^Jh.h#h. 2z { kBDVh$$Ifa$gd.gd.gd.gd.D8hjn$$Ifa$gd.gkd$$If40 0a$$Ifa$gd.gkdz$$If40 0a$$Ifa$gd.gkd$$If40 0a
N
P
p
r
t
v
úԩzúoSFooj
h#h.EHU6jDF
h.B*OJPJQJUV^JnH phtH jh#h.U*jh#h.B*CJU^JaJph0j.F
h#h.B*CJUV^JaJph!h#h.B*CJ^JaJphh#h.aJh#h.h#h.5\*jh#h.B*CJU^JaJph*jh#h.B*CJU^JaJph
$$Ifa$gd.gkdv$$If40 0a
`abcdef0^`0gdf/gd.gkd $$If40 0a
8DbeLNnprtv|~.2:>T\dhxԿԿԿԷԷԩԓxgggg h*h.CJEH
aJmHsHh*h.6]mHsHh#h.OJQJ^Jh*h.mHsHjh#h.UjF
h#h.UVh#h.H*h.h#h.CJEHaJhf/h#h.jh#h.Ujh#h.Uj0F
h#h.UV(-./01ghijk~0^`0gdf/gd.bcdeflu"JyzBC\
DLv`gd.0^`0gd.gd.gd.xz42464t4x4444444444444550565L5N5T5V5t5v5555568C8D8нh.h#h.CJEH
aJUh#h. h*h.CJEH
aJmHsHh*h.mHsHh#h.OJQJ^J(J444\5558C8D8gd.gd. moles = 1000 g / 18 gmol 1 = 56 mol
6. 56 mol 6.02 1023 mol 1 = 3 1025
7. (3.4 1025)/7000= 4.9 1021
8. energy released = 4.9 1021 (5.23 10 13 J) = 2.49 109 J
9. (2.49 109 J)/ (4200 J kg 1K 1 100K) = 6000 kg = 6000 litres
External reference
This activity is taken from Advancing Physics chapter 18, 260S
,1h. A!"#$%$$If!vh55#v:V40,5/4aDd
S0
#A2^Wڵazb# <6:X`!2Wڵazb# <6 Gxcdd`` @c112BYL%bL0Yn&B@?6 75bnĒʂT+~3v,L
E9@v2Br(T0zTBS<3>4VBa|.o>[;FLLJ%|iAh`J4$$If!vh55#v:V40,5/4aDd
?S0
#A2a*'3%m7=VX`!5*'3%m7֬ PGxcdd`` @c112BYL%bL0Yn&&&! KA?H17T
obIFHeA*C;a&Ǣ {7Br(To2zTBS[;FLLJ%ji] @4
nFf~7K'$$If!vh55#v:V40,5/4aDd
S0
#A2b ӧG/M>RX`!6 ӧG/M HXJxcdd`` @c112BYL%bL0Yn&B@?6 75bnĒʂT+~3v,L
E9@R.o >PNHq50ciͫy!#c~`kX _`4
*l
/`p121)W2'=cgbal_ v3209LT$$If!vh55#v:V40,5/4aDd
hhb
c$A??3"`?2Xs]-n5ހ7W4
X`!,s]-n5ހ7W@@||xcdd`` @c112BYL%bpu[;FLLJ% "CD1,Ġt`YGDd
S0
#A2c4tlI]ΨWtE
?X`!74tlI]ΨWtE
Ҭ @CXJxcdd`` @c112BYL%bL0Yn&LB@?6 7j䆪aM,,He`7Ӂ`0LYĠXd[|_H_30p1t9WM?@ABRoot Entry F&r*
Data
WordDocument.:ObjectPool (|r&r_1190378285F(|r(|rOle
CompObjfObjInfo !#$%&')
FMicrosoft Equation 3.0DS EquationEquation.39q5$
,23
He
FMicrosoft Equation 3.0DS EquationEquation.39qEquation Native @_1190378286F(|r(|rOle
CompObj
fObjInfo
Equation Native <_1191068818F(|r(|rOle
c
,01
n
FMicrosoft Equation 3.0DS EquationEquation.39q Gl
,12
HCompObjfObjInfo
Equation Native <_1190378288F(|r(|rOle
CompObjfObjInfoEquation Native @
FMicrosoft Equation 3.0DS EquationEquation.39qc$
,23
He
FMicrosoft Equation 3.0DS EquationEquation.39q_1190378143F(|r(|rOle
CompObjfObjInfoEquation Native 1Table6SummaryInformation(DocumentSummaryInformation8"P5/
,12
H + ,12
H!,23
He + ,01
nOh+'0|
8D
P\dltTAP 525-3: Fusion in a kettleAP mxgxgxgNormal.dotFmxg)Ύ^Y7@WW1`ڢ<{QR'^yŎ\_2*u7A)~\ͷWj&W?<|G%gHǲŌZIҮU}-ս1a4чQaq^z04)ԭh*,._4%dyHwq[ޙVW{lO`9}2gMicrosoft Word 10.0@F#@:nr@1r7՜.+,0 hp
Institute Of Physics - Londonn
TAP 525-3: Fusion in a kettleTitleH`H.Normal CJOJPJQJ_HmH sH tH DA@DDefault Paragraph FontRi@RTable Normal4
l4a(k@(No Listhoh.TAP MainH1$7$8$H$+5B* CJOJPJQJ\^JaJphtH JoJ.TAP ParaxB*CJOJQJ^JphFoF.TAP Subx5B*OJQJ^Jph`o!`.TAP Sub Char05B*CJOJPJQJ^J_HmH phsH tH : 2z{ k%&'789:;^_`a9:;<=~
Q
#
]
e
E00000000000000h 0h 0l 0h 0h 0l 0h 0h 0l 0h 0h 0l 0h 0h 0l 0000x000000000 0000000000000000000000000000x0x000000000000x000000@'0@'0@(0@'0@'0@(0@'0@'0@'0@'0@'0@'0@'0@'0@'0@'0@'0@'0@(0@'0\~00
xD8h
D8
D8>OQy:::::BM180060."9*urn:schemas-microsoft-com:office:smarttagsplace
4
5
6
7
A
B
I
J
K
N
S
U
W
X
]
E` q
\
33333` `
f/.%&@