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

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:

$\frac{dH}{dt}$ 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 , $\frac{dH_{1}}{dt}=67J/s$

• Final rate of heat ,  $\frac{dH_{2}}{dt}$= unknown

• Initial temperature , $T_{1}=374K$

• Final temperature , $T_{2}=439K$

$\frac{H_1}{T_{1}^4}=\frac{H_2}{T_{2}^4}$

$H_{2}=67\times1.898$

$H_{2}=127.2$.

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