Question

2 points
A black body at 1100°C has
surroundings at 200°C. Find heat
loss per unit area by radiation.*
O
1.9866 * 10^4 W/m2
O
1.9866 * 10^2 W/m2
O
1.9866 * 10^5 W/m2
O 1.9866 * 10^3 W/m2​

Answer 1

Answer:

The black body at 1100°C has its surroundings at 200°C, the heat loss per unit area by radiation will be 1.986 × 10⁵W/m².

Explanation:

The heat loss per unit area by radiation can be given by Stefan-Boltzmann law. The law relates the temperature to the amount of power it emits per unit area. The law can be stated as-

"The total energy emitted or radiated per unit surface area of a blackbody across all wavelength per unit time is directly proportional to the fourth power of blackbody's temperature".i.e.

P$/$A=eσ(T⁴-T⁴o)           (1)

P= energy radiated per unit time. Its unit is the watt(W).

A=area

e=emissivity

σ=5.67×10⁻⁸W$/$m²K⁴     (2)

T⁴= fourth power of black body temperature in kelvin(K)

T⁴o= fourth power of surrounding temperature in kelvin(K)

P$/$A can be written as U  (3)

Now as per the question,

T=1100°C +273=1372K  (4)

To=200°C+273=473K  (5)

And we will take e=1 because we are considering an ideal case of radiation.  (6)

By using equations (1),(2),(3),(4),(5), and (6) we get;

U=1×5.67×10⁻⁸×[(1373)⁴-(473)⁴]

U=5.67×10⁻⁸×3.503×10¹²

U=1.986 × 10⁵W/m²

Hence for the black body at 1100°C has its surroundings at 200°C, the heat loss per unit area by radiation will be 1.986 × 10⁵W/m².