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Chapter 6: Q60P (page 1317)

A large cavity that has a very small hole and is maintained at a temperature of T is a good approximation to an ideal radiator or blackbody. Radiation can pass into or out of the cavity only through the hole. The cavity is a perfect absorber since any radiation incident on the hole becomes trapped inside the cavity. Such a cavity at 400°C has a hole with area 4.00mm2. How long does it take for the cavity to radiate 100 J of energy through the hole?

Short Answer

Expert verified

The time taken by the cavity radiating 100 J of energy through the hole is 2.45h.

Step by step solution

01

Stefan-Boltzmann law

The Stefan-Boltzmann law states that the total radiant heat power emitted from a surface is proportional to its absolute temperature to the fourth power.

l=σT4

02

Time taken by the cavity radiates 100 J of energy through the hole

From the definition of radiation intensity;

l=PA

Derived from the definition of power;

P=Et

From the law of Stefan Boltzmann

l=σT4

Therefore;

l=PA=EAt=σT4

Calculating time T;

t=EσT4At=100j5.6707×10-8W/m2.K4200+273.15K4×10-6m2=8800s=2.45h

Hence, the time taken by the cavity radiating 100 J of energy through the hole is 2.45h.

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