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Temperature of substances A and B are 72...

Temperature of substances A and B are `727^(@)C` and `327^(@)C` respectively, then ratio of rate of their radiant energy `H_(A):H_(B)=` . . . . .

A

`727:327`

B

`5:3`

C

`25:9`

D

`625:81`

Text Solution

Verified by Experts

The correct Answer is:
D

`T_(A)=273+727=1000K`
`T_(B)=273+327=600K`
Rate of radiant energy
`HpropT^(4)`
`:.(H_(A))/(H_(B))=(T_(A)^(4))/(T_(B)^(4))`
`=((1000)^(4))/((600)^(4))`
`=((10)/(6))^(4)`
`=((5)/(3))^(2)`
`=(625)/(81)`
`:.H_(A):H_(B)=625:81`
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