SME
Grade 12 Science · Problem type
Earth’s energy balance
Use this for “If Earth reflects 30% of sunlight and has no greenhouse gases, how warm is it?”
Diagram
Averaged
over
the
globe,
1,361
4
=
340
W/
m
2
arrives.
With
α
=
0.3,
102
is
reflected
and
F
=
1,361
×
1 − 0.3
4
=
238
W/
m
2
is
absorbed.
In balance the ground sends F back out: σTₑ⁴ = 238, so Tₑ = 255 K (−18 °C).
This is an example. Type your own values to start a new problem.
5.67
×
10⁻⁸
×
(
Tₑ (K)
)
4
=
S (W/m²)
×
(1
−
α
)
÷
4
=
F (W/m²)
Formulas
F
=
S
×
1 −
α
4
238
=
1,361
×
1 − 0.3
4
Tₑ
=
∜(
F
5.67 ×
10
-8
)
255
=
∜(
238
5.67 ×
10
-8
)
Assumptions
•
Sunlight falls on Earth’s disk but spreads over the whole globe, 4 times the disk’s area, so each square meter gets S ÷ 4 on average.
•
The albedo α is the share reflected by clouds, ice and land; the rest is absorbed.
•
In balance the ground sends out as infrared what it absorbs: σTₑ⁴ = F, with σ = 5.67 × 10⁻⁸ W/m² per K⁴.
Step-by-step
WE KNOW
S = 1,361 W/m², α = 0.3
FIND
sunlight absorbed (F), balance temperature (Tₑ)
Step 1 · Find sunlight absorbed (F)
Use F = S × (1 − α) ÷ 4
The share not reflected, spread over 4 times the disk’s area.
F
=
1,361
×
1 − 0.3
4
F
=
1,361
×
0.7
4
F
=
952.7
4
F
=
238
W/
m
2
Step 2 · Find balance temperature (Tₑ)
Use Tₑ = ∜(F ÷ (5.67 × 10⁻⁸))
The
temperature
whose
infrared,
σTₑ⁴,
carries
away
F
.
Tₑ
=
∜(
238
5.67 ×
10
-8
)
Tₑ
=
∜(4.2
×
10
9
)
Tₑ = 255 K (about −18 °C; Earth averages about 15 °C, so greenhouse gases add about 33 °C)
Check
Put the numbers back in. Both sides should match.
238
≈
1,361
×
1 − 0.3
4
✓
255
≈
∜(
238
5.67 ×
10
-8
)
✓
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