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How are Stefan's law and Newton's law of cooling related?

seo-qna
Last updated date: 13th Jun 2024
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Answer
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Hint: The rate of cooling of a body is directly proportional to the difference in temperature between the body and the surrounding area, according to Newton's cooling, provided that the temperature difference is very small. Study each law given in the alternatives and which fits Newton's law.
Formula used:
$\dfrac{d Q}{d t}=e \sigma\left(T^{4}-T_{0}^{4}\right)$

Complete answer:
The rate of cooling of a body is directly proportional to the difference in temperature between the body and the surrounding area, according to Newton's cooling, provided that the temperature difference is very small.
i.e. $\dfrac{d Q}{d t} \propto\left(\theta-\theta_{0}\right)$
Here, $\theta$ is the temperature of the cooling body and $\theta_{0}$ is the temperature of the surrounding.
Newton's law cooling is a special case of Stefan-Boltzmann's law where the temperature difference of the body and the surrounding is very small.
Let's prove the above statement.
According to Stefan-Boltzmann's law, the rate of cooling of a body is given as $\dfrac{d \mathrm{Q}}{d t}=e \sigma\left(T^{4}-T_{0}^{4}\right)$..... (ii).
Here, e is the emissivity of the body, $\sigma$ is Stefan-Boltzmann's constant, $T$ is the temperature of the cooling body and $T_{0}$
is the temperature of the surrounding.
Suppose the temperature difference of the body and the surrounding is $\Delta T=T-T_{0}$.
$\Rightarrow T=T_{0}+\Delta T$
Substitute the value of $T$ in equation (ii).
$\Rightarrow \dfrac{d Q}{d t}=e \sigma\left(\left(T_{0}+\Delta T\right)^{4}-T_{0}^{4}\right)$
$\Rightarrow \dfrac{d Q}{d t}=e \sigma\left(T_{0}^{4}\left(1+\dfrac{\Delta T}{T_{0}}\right)^{4}-T_{0}^{4}\right) \ldots . .$ (iii).
When we have a term $(1+x)^{4}, x$ is a very small number (close to zero), the term is approximately equal to $(1+n x)$.
i.e. $(1+x)^{n} \approx 1+n x$.
Consider the term $\left(1+\dfrac{\Delta T}{T_{0}}\right)^{4}$ Since $\Delta T$ is very small, the ratio $\dfrac{\Delta T}{T_{0}}$ is also very small. Hence, $\left(1+\dfrac{\Delta T}{T_{0}}\right)^{4} \approx 1+4 \dfrac{\Delta T}{T_{0}}$
Substitute this value in (iii). $\Rightarrow \dfrac{d Q}{d t}=e \sigma\left(T_{0}^{4}\left(1+4 \dfrac{\Delta T}{T_{0}}\right)-T_{0}^{4}\right)$
$\Rightarrow \dfrac{d Q}{d t}=e \sigma\left(T_{0}^{4}+4 \Delta T T_{0}^{3}-T_{0}^{4}\right)$
$\Rightarrow \dfrac{d Q}{d t}=e \sigma\left(4 \Delta T T_{0}^{3}\right)$
$\Rightarrow \dfrac{d Q}{d t}=e \sigma 4 T_{0}^{3}\left(T-T_{0}\right)$
Since $e \sigma 4 T_{0}^{3}$ is a constant value, $\dfrac{d Q}{d t} \propto\left(T-T_{0}\right)$
Hence, proved that the Stefan-Boltzmann's law is the same as that of Newton's law of cooling for small temperature difference.

Note:
Let us understand what we are told by the other laws given in the options.
(I) Kirchhoff’s law: it states that, for all surfaces, the ratio of emissive power to absorptive power is the same on the same surface. Temperature and is equal at that temperature to the emissive power of a perfectly black body.
(ii) Wien's law: According to this law, the wavelength product for which a black body emits maximum intensity radiation and the body temperature is constant.