Blackbody radiation calculation problem

Click For Summary
SUMMARY

The discussion centers on calculating the energy radiated by an average person using the Stefan-Boltzmann law. The formula used is L = AσT^4, where A is the surface area (1.4 m²), σ is the Stefan-Boltzmann constant (5.670400 × 10^-8 W/m²K⁴), and T is the temperature in Kelvin. The correct temperature for radiation is 306 K, while the absorbed temperature is 293 K. The energy radiated per second is calculated as L = (1.4 m²)(5.670400 × 10^-8 W/m²K⁴)(306 K)^4.

PREREQUISITES
  • Understanding of the Stefan-Boltzmann law
  • Knowledge of temperature scales, specifically Kelvin
  • Familiarity with basic physics concepts related to heat transfer
  • Ability to perform calculations involving exponents and scientific notation
NEXT STEPS
  • Study the derivation and applications of the Stefan-Boltzmann law
  • Learn about blackbody radiation and its significance in thermodynamics
  • Explore the concept of emissivity and its effect on energy radiation
  • Investigate heat transfer mechanisms in different environments
USEFUL FOR

Students in physics, engineers working with thermal systems, and anyone interested in understanding heat radiation principles.

LordCalculus
Messages
12
Reaction score
0
Blackbody radiation problem!

Homework Statement


The average person has 1.4 m^2 of skin at a skin temperature of roughly 306 K. Consider the average person to be an ideal radiator standing in a room at a temperature of 293 K.

a.) Calculate the energy per second radiated by the average person in the form of blackbody radiation. Express you answer in watts. (How do I treat the temperature?)

Homework Equations



L=AσT^4

Where A = area; σ = Stefan-Boltzmann constant; T = temperature (in Kelvins).

The Attempt at a Solution



L=(1.4)*(5.670400*10^-8)*(?)^4
 
Physics news on Phys.org


The energy per second radiated will just be that equation with body temperature as T (~310 Kelvin); but note that this isn't the same as the total energy lost per second.
 


Energy radiated per unit time will be given by Stefan-Boltzmann law -

dQ/dt = eAσT^4 (σ - Stefan's constant)

but here 'T' is to be taken in Kelvin and not degree celsius.

and also this is not the energy lost since energy lost = (Energy radiated)-(Energy absorbed)

put T value in the equation. For heat radiated T=306 and for absorbed T=293
 

Similar threads

Replies
6
Views
1K
  • · Replies 29 ·
Replies
29
Views
4K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
966
  • · Replies 4 ·
Replies
4
Views
3K
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
14
Views
5K
  • · Replies 1 ·
Replies
1
Views
3K