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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

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Answer:

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

Explanation:

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

[tex]\frac{dH}{dt}[/tex] is directly proprtional to the  [tex]T^{4}[/tex].

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 , [tex]\frac{dH_{1}}{dt}=67J/s[/tex]
  • Final rate of heat ,  [tex]\frac{dH_{2}}{dt}[/tex]= unknown
  • Initial temperature , [tex]T_{1}=374K[/tex]
  • Final temperature , [tex]T_{2}=439K[/tex]

[tex]\frac{H_1}{T_{1}^4}=\frac{H_2}{T_{2}^4}[/tex]

[tex]H_{2}=67\times1.898[/tex]

[tex]H_{2}=127.2[/tex].

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

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