Question

How many days does it take for a perfect blackbody cube (0.0134 m on a side, 35.7 °C) to radiate the same amount of energy that a one-hundred-watt light bulb uses in one hour?

Answer 1

power radiated by blackbody =sigma*A*T^4

where sigma=stefan's constant =5.67*10^(-8)

A=total surface area=6*0.0134^2=0.0010774 m^2

T=temperature in kelvin=273+35.7=308.7 K

if time taken to match the energy delivered by the bulb is t seconds

then power*t=power of bulb*1 hour

==>5.67*10^(-8)*0.0010774*308.7^4*t=100 W *3600 seconds

==>t=100*3600/(5.67*10^(-8)*0.0010774*308.7^4)=6.4893*10^5 seconds

=6.4893*10^5/(3600 (seconds/hour)*24 (hours/day))=7.5107 days