SME
Geophysics · Problem type
Seafloor heat flow from its age
Use this for “Seafloor is 50 million years old, with k = 3.3 W/(m·K), κ = 1 mm²/s and a mantle 1,300 K hotter than the sea floor. What heat flow comes out of it?”
Chart
q(50) ≈ 60.93
This is an example. Type your own values to start a new problem.
q (mW/m²)
=
k (W/(m·K))
×
ΔT (K)
÷
π
×
κ (mm²/s)
×
t (Myr)
Formulas
q
=
k
×
ΔT
÷
√
π ×
κ
× 3.156 ×
10
13
×
t
0.06093
=
3.3
×
1,300
÷
√(π
×
(1
×
10
-6
)
×
3.156 ×
10
13
×
50)
Formulas are worked in m²/s, W/m². The step-by-step shows the conversions.
Assumptions
•
The plate cools from the top like a half-space: the closed form q = kΔT ÷ √(πκt) stands in for solving the cooling equation.
•
ΔT is how much hotter the mantle starts than the sea floor; one million years is 3.156 × 10¹³ s, so the age is turned into seconds first.
•
The fit holds for young floor; past about 80 Myr the heat flow levels off near 50 mW/m².
Step-by-step
WE KNOW
t = 50 Myr, k = 3.3 W/(m·K), ΔT = 1,300 K, κ = 1 mm²/s
FIND
heat flow (q)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to m²/s, W/m²).
κ = 1 mm²/s = 1 × 10⁻⁶ m²/s (1 m²/s = 10⁶ mm²/s)
Step 2 · Find heat flow (q)
Use q = k × ΔT ÷ √(π × κ × 3.156 × 10¹³ × t)
Conductivity
times
the
temperature
drop,
over
the
depth
the
cooling
has
reached,
√
πκt
.
q
=
3.3
×
1,300
÷
√
π ×
10
-6
× 3.156 ×
10
13
× 50
50
×
3.156 ×
10
13
=
1.578 ×
10
15
π
×
10
-6
×
1.578 ×
10
15
=
4.957 ×
10
9
√
4.957 ×
10
9
=
70,409
m
3.3 × 1,300 = 4,290
q
=
0.06093
W/
m
2
Step 3 · Convert the answers
Change each answer back to the unit you chose.
q = 0.06093 W/m² = 60.9297 mW/m² (1 W/m² = 1000 mW/m²)
Check
Put the numbers back in. Both sides should match.
0.06093
=
3.3
×
1,300
÷
√
π ×
10
-6
× 3.156 ×
10
13
× 50
✓
Topic check
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