Question

The rate of cooling at 600k, if surrounding temperature is 300K, is H. The rate of cooling at 900K is-
A. $\dfrac{{16}}{3}H$
B. $2H$
C. $3H$
D. $\dfrac{2}{3}H$

Answer 1

Hint: We will first apply the Stefan’s law and the formula $R = \varepsilon \sigma ({T^4} - {T_0}^4)$ Where R is the rate of cooling and ${T_0}$ is the surrounding temperature. Then we will divide the two equations we will get by applying this law in order to get our desired answer. Refer to the solution below.
Formula used: $R = \varepsilon \sigma ({T^4} - {T_0}^4)$

Complete Step-by-Step solution:
As we know that the rate of cooling is directly proportional to the difference of the quadruple of the temperatures-
(According to Stefan’s law)
Rate of cooling $ \propto \left( {{T^4} - {T_0}^4} \right)$
Let the rate of cooling at 900K be H’.
T = 900K (given)
And ${T_0}$ (surrounding temperature) = 300K (given)
Applying these values in the above law, we get-
$ \Rightarrow H' = \varepsilon \sigma \left( {{{900}^4} - {{300}^4}} \right)$
Let the above equation be equation 1, we get-
$ \Rightarrow H' = \varepsilon \sigma \left( {{{900}^4} - {{300}^4}} \right)$ (equation 1)
The rate of cooling at 600K as given in the question is H. So-
$ \Rightarrow H = \varepsilon \sigma \left( {{{600}^4} - {{300}^4}} \right)$
Let the above equation be equation 2, we get-
$ \Rightarrow H = \varepsilon \sigma \left( {{{600}^4} - {{300}^4}} \right)$ (equation 2)
Now, diving equation 1 with equation 2, we get-
$
   \Rightarrow \dfrac{{H'}}{H} = \dfrac{{\varepsilon \sigma \left( {{{900}^4} - {{300}^4}} \right)}}{{\varepsilon \sigma \left( {{{600}^4} - {{300}^4}} \right)}} \\
    \\
   \Rightarrow \dfrac{{H'}}{H} = \dfrac{{\left( {{{900}^4} - {{300}^4}} \right)}}{{\left( {{{600}^4} - {{300}^4}} \right)}} \\
    \\
   \Rightarrow \dfrac{{H'}}{H} = \dfrac{{81 - 1}}{{16 - 1}} \\
    \\
   \Rightarrow \dfrac{{H'}}{H} = \dfrac{{80}}{{15}} \\
    \\
   \Rightarrow \dfrac{{H'}}{H} = \dfrac{{16}}{3} \\
    \\
   \Rightarrow H' = \dfrac{{16}}{3}H \\
$
Thus, option A is the correct option.

Note: According to the law of Stefan Boltzmann the level of radiation released per unit of time from a black body region A at complete temperature T is directly proportional to the temperature's fourth power. “The total energy emitted / radiated by a black body's unit of surface area over all wavelengths per time unit is directly proportional to the fourth power of thermodynamic temperature of the black body.”