Question

A cube at 101 °C radiates heat at a rate of 67 J/s. If its surface temperature is increased to 166 °C, the rate at which it will now radiate is closest to

Answer 1

The rate at which it will radiate heat is closest to 127.2J.

Step-by-step explanation:

According to Stefan's law, the heat radiated by a black body can be written as:

`(dH)/(dt)` is directly proprtional to the
`T^(4)`.

where H is the amount of heat radiated and T is the surface temperature .

Example : The heat is transferred by radiation and an example of heat transferred by radiation is the sunlight reaching earth from sun.

Note : Here the temperature is expressed in Kelvin.

• Initial rate of heat ,
  `(dH_(1))/(dt)=67J/s`

• Final rate of heat ,
  `(dH_(2))/(dt)`= unknown

• Initial temperature ,
  `T_(1)=374K`

• Final temperature ,
  `T_(2)=439K`

`(H_1)/(T_(1)^4)=(H_2)/(T_(2)^4)`

`H_(2)=67*1.898`

`H_(2)=127.2`.

The rate at which it will radiate heat is closest to 127.2J.