SME
Heat Transfer · Problem type
A blackbody’s peak wavelength and emissive power
Use this for “The Sun’s surface is close to a 5800 K blackbody. At what wavelength does it radiate most, and what is its emissive power?”
Chart
5,800
K:
λₘₐₓ
=
b
T
=
2,898
μm·
K
5,800
K
=
0.4997
μm
Tinted: the visible band, 380 to 750 nm
This is an example. Type your own values to start a new problem.
λₘₐₓ (μm)
×
T (K)
=
2898
μm·K
Eb (W/m²)
=
5.67
×
10
−8
×
T⁴
Formulas
λₘₐₓ
×
T
=
2,898
0.4997 × 5,800 = 2,898
Eb
=
5.67 ×
10
-8
×
T
4
6.4165 ×
10
7
=
5.67 ×
10
-8
×
5,
800
4
Assumptions
•
A blackbody (ε = 1): it absorbs all radiation and emits the most possible at its temperature.
•
Wien’s law: λₘₐₓ T = 2898 μm·K; T in kelvins.
•
Stefan–Boltzmann: Eb = σT⁴, σ = 5.67 × 10⁻⁸ W/(m²·K⁴); a real surface emits εEb.
Step-by-step
WE KNOW
T = 5,800 K
FIND
peak wavelength (λₘₐₓ), emissive power (Eb)
Step 1 · Find peak wavelength (λₘₐₓ)
Use λₘₐₓ × T = 2,898
Wien’s law: the hotter the body, the shorter the wavelength where its spectrum peaks.
λₘₐₓ
=
2,898
T
λₘₐₓ
=
2,898
5,800
λₘₐₓ = 0.4997 μm
Step 2 · Find emissive power (Eb)
Use Eb = 5.67 × 10⁻⁸ × T⁴
Stefan–Boltzmann: the whole spectrum summed over every wavelength gives σT⁴ per square metre.
Eb
=
5.67 ×
10
-8
×
5,
800
4
Eb
=
(5.67
×
10
-8
)
×
(1.1316
×
10
15
)
Eb
=
6.4165 ×
10
7
W/
m
2
Check
Put the numbers back in. Both sides should match.
0.4997 × 5,800 = 2,898 ✓
6.4165 ×
10
7
=
5.67 ×
10
-8
×
5,
800
4
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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