Home
Class 11
PHYSICS
The temperature of a perfect black body ...

The temperature of a perfect black body is 1 000K and its area is 0.1 `m^(2)` . If `sigma = 5.67 xx 10^(-8) ` W `m^(-2) K^(-4)` calculate the heat radiated by it in 1 minute ?

Text Solution

Verified by Experts

According to Stefan.s law if Eis the energy radiated in 1 sec by unit surf ace area then E `sigma T^(4)`
`therefore` Total energy radiated , E = A `(sigma T^(4))`
`t = 0.1 xx 5.67 xx 10^(-8) xx 1000^(4) xx 60`
= `34.02 xx 10^(4)` J
Promotional Banner

Similar Questions

Explore conceptually related problems

The power of a black body at temperature 200K is 544 watt. Its surface area is (sigma = 5.67 xx 10^(-8) wm^(-2) K^(-4))

Heat loss takes place from a body maintained at a temperature of 400^(@)C to the surrounding air at 30^(@)C by convection and to the surrounding surfaces at 30^(@)C by radiation. The Newton's cooling coefficient is 20 W//m^(2) K and the Stefan-Boltzmann constant is 5.67 xx 10^(-8) W//m^(2)K^(4) . If the rate of heat loss by convection is equal to the rate of heat loss by radiation, the emissivity of the body surface is