A black body, at a temperature of `227^@C`, radiates heat at a rate of 20 cal `m^(-2)s^(-1)`. When its temperature is raised to `727^@C`, the heat radiated by it in cal `M^(-2)s^(-1)` will be closest to

Options

A. 40
B. 160
C. 320
D. 640

Answer

C

Explanation

The temperature of the black body is
`T_1=227^@C=500K`
`therefore` Using Stefan's law , the rate of heat radiation per unit area per unit time is
`E_1= sigmaT^4 rArr 20 = sigma(500)^4 rArr sigma = (20)/((500)^4)`
Now the temperature of the blackbody id raised to
`T_2=272^@C=1000K`
`therefore` Rate of heat radiation per unit area
`E_2=sigmaT_2^4=(20)/((500)^4)xx(1000)^4`
`=20xx2^4 = 320 cal m^(-2) s^(-1)`