Question

The rate of cooling at $600K$, if the surrounding temperature is $300K$, is $H$. The rate of cooling at $900K$ is

• A. $\frac{16}{3}H$
• B. $2H$
• C. $3H$
• D. $\frac{2}{3}H$

Answer 1

The correct option is A

$\frac{16}{3}H$

Step 1: Given data:

The rate of cooling is $H$ at temperature $T=600K$ with the surrounding temperature $T_0=300K$

To find: Rate of cooling is $H\text{'}=?$ at temperature $T\text{'}=900K$ with surrounding temperature $T_0=300K$

Step 2: Formula used:

According to Stefan's law, the rate of cooling $H\propto \left(T^4-{T_0}^4\right)$, where $T$ is the given temperature and $T_0$ is the surrounding temperature.

Step 3: Calculation of cooling rate at $900K$ :

$$\begin{gathered} H\text{'}\propto \left(T{\text{'}}^4-{T_0}^4\right) \\ H\propto \left(T^4-{T_0}^4\right) \end{gathered}$$

Taking ratio of both above equations,

$$\begin{gathered} \frac{H\text{'}}{H}=\frac{T{\text{'}}^4-{T_0}^4}{T^4-{T_0}^4} \\ \frac{H\text{'}}{H}=\frac{{900}^4-{300}^4}{{600}^4-{300}^4} \\ \frac{H\text{'}}{H}=\frac{9^4-3^4}{6^4-3^4} \\ \frac{H\text{'}}{H}=\frac{3^4\left(3^4-1\right)}{3^4\left(2^4-1\right)} \\ \frac{H\text{'}}{H}=\frac{81-1}{16-1} \\ \frac{H\text{'}}{H}=\frac{80}{15} \\ \frac{H\text{'}}{H}=\frac{16}{3} \\ H\text{'}=\frac{16}{3}H \end{gathered}$$

Thus, the rate of cooling will be $\frac{16}{3}H$.

So, option A is correct.