The Solid Earth Chapter 7 Answers to selected … › instruction › HSU › 2016_spring › ...The assumptions are that Venus has an Earth-like isotopic composition, an Earth-like

The Solid EarthChapter 7

Answers to selected questions

(1) At 4000 Ma, abundances 238U, 235U, Th, K: 1.85, 48.6, 1.22, 8.37. At 4500 Ma, abundances 238U, 235U, Th, K: 1.99, 78.9, 1.25, 10.9.

(2) Use Eq. 7.42.Daily at z=2 m:

φ =7.27 ×10−5 × 2.3×103 ×103

2 × 2.5× 2

= 11.6Daily at z=5 m:

φ =7.27 ×10−5 × 2.3×103 ×103

2 × 2.5× 5

= 28.9 Phase difference is -17.3 radians.

Annual at z=2 m:

φ =2 ×10−7 × 2.3×103 ×103

2 × 2.5× 2

= 0.61Annual at z=5 m:

φ =2 ×10−7 × 2.3×103 ×103

2 × 2.5× 5

= 1.52 Phase difference is -0.91 radians.

(3) Use Eq. 7.45:

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5 = 40exp −ωρcp

2kz

exp i ωt −

ωρcp2k

z

0.125 = exp 0.092z( )exp i ωt − 0.092z( )[ ]0.125 = exp −0.3033z( ) gives the amplitude−2.079 = −0.3033z

z = 6.9 m

Use Eq. 7.42 to calculate the phase:

φ =ωρcp

2kz

= 0.3033× 6.9= 2.09 radians =17 weeks

(4) T =Tddz

(5) Parameters for this problem:

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dTdz

= 30 at z = 0

T = 700 at z = 25 km. A =1×10−6 Wm-3

k = 3 Wm-1K-1

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d2Tdz2 = −

Ak

dTdz

= −Akz + c1 At z = 0, c1 = 30 ×10−3

T = −A2k

z2 + c1z + c2

T = −10−6

6z2 + 3×10−2 z + c2

To evaluate c2 use T = 700 at z = 35 km.

700 = −35 × 35

6+ 3×10−2 × 35 ×103 + c2

c2 = 700 + 204 −1050 = −146

The geotherm is therefore: T = −z2

6+ 30z −146

(6) Venera 8For K: 1 m3 of the sample contains

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0.47 ± 0.08( ) × 2800100

=13.16 ± 2.24 kg of K

40K is 0.012% of total K so the heat production from K is:

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13.16 ± 2.24( ) × 0.012100

× 2.8 ×10−5 = 0.044 ± 0.007( ) ×10−6 Wm-3

= 0.044 ± 0.007 µWm-3

Heat production from U is

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0.6 ± 0.16( )106 × 2800 × 0.9928 × 9.4 ×10−5 + 0.0072 × 5.7 ×10−4( ) = 0.16 ± 0.04( ) ×10−6 Wm-3

= 0.16 ± 0.04 µWm-3

Heat production from Th is

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3.65 ± 0.42( )106 × 2800 × 2.7 ×10−5 = 0.28 ± 0.03( ) ×10−6 Wm-3

= 0.28 ± 0.03 µWm-3

Total heat production is

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0.48 ± 0.08( ) ×10−6 Wm-3 = 0.48 ± 0.08 µWm-3

For Venera 9, total heat production is 0.21±0.11 µWm-3

For Venera 10, total heat production is 1.47±0.31 µWm-3

(After Nisbet and Fowler, 1982).

(7) Use Eq. 7.27.For Venera 8:

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T = −0.096z2 +18z + 740, z in km, T in KT = −0.096z2 +18z + 467, z in km, T in oC

For Venera 9:

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T = −0.042z2 +12.6z + 740, z in km, T in KT = −0.042z2 +12.6z + 467, z in km, T in oC

For Venera 10:

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T = −0.294z2 + 37.8z + 740, z in km, T in KT = −0.294z2 + 37.8z + 467, z in km, T in oC

The assumptions are that Venus has an Earth-like isotopic composition, anEarth-like heat flow from depth, that the heat production is uniform in theoutermost 50 km and that an equilibrium geotherm is valid.The Al2SiO5 triple point is shown as a dashed line. It is clear that high-pressure low-temperature assemblages do not exist on Venus. It is alsoapparent that since neither water nor mud exist on Venus, neither do most ofthe classical metamorphic minerals found on Earth. Students may also beencouraged to think about controls on partial; melting in dry settings. Thelikelihood of diamond mining on Venus could also be speculated upon.(After Nisbet and Fowler, 1982).

(8) O1 geotherm:T = −0.134z2 +10.5z for 0 ≤ z<10 kmT = 7.84z +13 for z ≥ 10 km

O2 geotherm:T = −0.067z2 +13.4z for 0 ≤ z<10 kmT = 12.05z + 7 for z ≥ 10 km

C1 geotherm:

T = 90 1− e− z 11( ) + 6.5z for 0 ≤ z<40 kmT = 6.7z + 79 for z ≥ 40 km

C2 geotherm:

T = 49 1− e− z 11.5( ) +10.4z for 0 ≤ z<40 kmT = 10.5z + 44 for z ≥ 40 km

(9) Use Eq. 7.31-7.32 with A1=42 µWm-3, A2=4.8 µWm-3, z1=20 km, z2=35 km, Qd=63 mWm-2,

k=2.52 Wm-1K-1.T = −0.672 + 50.7z for 0 ≤ z<20 kmT = −0.127z2 + 29z + 217 for 20 ≤ z<35 km

(10) Use Eq. 7.46 to calculate the skin depth for an ice age with period 105 years:

ω =

2π105 × 365 × 24 × 60 × 60

= 2 ×10−12 s-1

skin depth = 2 × 2.52 ×10−12 × 2.3×103 ×103

=2.52.3

×106

= 1.04 ×103 metres

(11) Use Eqs 7.31-7.32 with: A1=2.5 µWm-3, A2=0, z1=10 km, z2=35 km, Qd=20 mWm-2, k=2.5 Wm-1K-1.

T = −0.5z2 +18z for 0 ≤ z<10 kmT = 8z + 50 for 10 ≤ z<35 km

(12) Use Eqs 7.31-7.32 with: A1=0, A2=1 µWm-3, z1=10 km, z2=35 km, Qd=20 mWm-2, k=2.5 Wm-1K-1.

T = 18z for 0 ≤ z<10 kmT == 0.2z2 + 22z − 20 for 10 ≤ z<35 km

Temperature gradient is linear through layers with zero heat generationand parabolic through layers with heat generation. Higher temperatures occurwith heat generation at depth.

(13) (a) Use Eqs 7.31-7.32 with: A1=2.5 µWm-3, A2=0.4 µWm-3, z1=10 km, z2=40 km, Qd=20 mWm-2,

k=2.5 Wm-1K-1.T = −0.5z2 + 22.8z for 0 ≤ z<10 kmT = −0.08z2 +14.4z + 42 for 10 ≤ z<40 km

(b) Use Eqs 7.31-7.32 with:A1=0.4 µWm-3, A2=2.5 µWm-3, z1=30 km, z2=40 km, Qd=20 mWm-2,k=2.5 Wm-1K-1.T = −0.08z2 + 22.8z for 0 ≤ z<30 kmT = −0.5z2 + 48z − 378 for 10 ≤ z<40 km

(14) (a) Intrusion width is 1m, so ω is 0.5m. Diffusivity, κ is 10-6 Nm-2. Since theintrusion temperature is 1050°C and the country rock temperature is 50°C, useEq. 7.37 with To=1000°C and a solidus temperature of 750°C:

750 = 1000 erf 0.52 10−6 t

0.75 = erf 0.52 10−6 t

Use Table A5.1 to determine the error function:

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10−6 t = 0.094t = 9.4 ×104 seconds

= 26 hours(c) Use Eq. 7.37 to calculate the temperature at 2 weeks:

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t = 2 × 7 × 24 × 60 × 60=1.2 ×106 seconds

T =1000erf 0.52 10−61.2 ×106

=1000erf 0.228[ ]= 253 oC

So after 2 weeks the temperature at the centre of the dyke will be 253°C.

(15) Assume that the basalt is erupted onto the surface into a medium at 0°C. Use Eq. 7.37:

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900 =1200erf 103

2 10−6 t

0.75 = erf 103

2 10−6 t

0.814 =103

2 10−6 t10−6 t = 614

t = 3.8 ×1011 seconds=1.2 ×104 years

Contact zone, extent of metamorphism: 250 m from the basalt. At thisdistance from the basalt the highest temperature reached would be ca300°C. Temperatures of 600°C would be reached only within 125 m ofthe basalt. Latent heat is an important additional heat source that is notincluded in this solution.

(16) (a) Age of lithosphere 200 km from the ridge axis:200 ×103

2.5 ×10−2 = 8 ×106 years

Use Eq. 7.57a to determine the bathymetric depth for 8Ma old lithosphere:

d = 2.6 + 0.365 8= 3.63kmDifference in depth at intersection is 1.03 km.

(b) At 1000 km from the ridge, the age of the lithosphere is 40 Ma.Use Eq. 7.57b:Depth is 4.84 km at 40 Ma and 5.00 km at 48 Ma.The difference in depth is ca 160 m.

At 3000 km from the ridge, the age of the lithosphere is 120 Ma.Use Eq. 7.57b:Depth is 5.56 km at 120 Ma and 5.58 km at 128 Ma.The difference in depth is ca 20 m.

(17) To calculate the 60 Ma geotherm use Eq. 7.61:

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T z( ) =1300erf z2 10−6 × 60 ×106 × 3.16 ×107

=1300erf z8.7 ×104

The base of the lithosphere will be at that depth, L, where T(z)=1150°C:

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1150 =1300erf L8.7 ×104

0.8846 = erf L8.7 ×104

1.115 =L

8.7 ×104 using Table A5.1 for the error function

L = 9.7 ×104 metres= 97 km

(18) Use Eq. 7.56. Surface heat flow is 87 mWm-2, so the heat generation A, uniformlydistributed in a conducting sphere needed to cause this is:

87 ×10−3 =A × 6370 ×103

3

A = 87 ×10−3 × 36370 ×103

= 0.041×10−6 Wm-3

= 0.041 µWm-3

The average heat generation of the crust is ca 0.5-0.7 µWm-3 and for themantle ca 0.02 µWm-3 (Table 7.1).

(19)

Volume of the crust = 10 ×103 × 4π × 6371×103( )2

= 5.1×1018 m3

Heat produced in the crust = 1.5 ×10−6 × 5.1×1018

= 7.65 ×1012 W

Volume of the mantle = 4π3

6361×103( )3− 3480 ×103( )3

= 9 ×1020 m3

Heat produced in the mantle = 1.5 ×10−8 × 9 ×1020

= 1.35 ×1013 W

(20) Use Eq. 7.56:

heat flow = 2 ×10−8 × 6371×103

3= 4247 ×10−5

= 42 ×10−3 Wm-3

(21) Hen’s egg, 0.06kg, ostrich egg, 1.4 kg. Assume both have the same density ρand are spherical, radius rh and ro.

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43πrh

3ρ = 0.06

43πro

3ρ =1.4

rh3

ro3 =

0.061.4

rh ro

= 0.35

The characteristic time is proportional to the square of the length scale, so asthe time needed to cook the hen egg is 4 minutes the time required to cook the

ostrich egg is 4

0.352 = 33 minutes

(22) (a) Assume κ=10-6 m2s-1. The characteristic time for the Earth is then

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6371×103( )2

10−6 = 4 ×1019 seconds

=1.27 ×106 Ma (b)

L2 = 4500 ×106 × 3.16 ×107 ×10−6

= 1.42 ×1011

L = 3.8 ×105 metres= 380 km

(c) These values clearly show that conduction cannot be the main heattransfer mechanism operating in the Earth – it is too slow.