Question

Newton’s law of cooling is a special case of
A. Kirchoff’s law
B. Wein’s law
C. Stefan-Boltzmann’s law
D. Planck’s law

Answer 1

Hint: According to Newton's cooling, the rate of cooling of a body is directly proportional to the difference of temperatures of the body and the surrounding, provided that the temperature difference is very small. Study each law given in the options and which Newton's law fits.

Formula used:
$\dfrac{dQ}{dt}=e\sigma \left( {{T}^{4}}-T_{0}^{4} \right)$

Complete answer:
According to Newton's cooling, the rate of cooling of a body is directly proportional to the difference of temperatures of the body and the surrounding, provided that the temperature difference is very small.
i.e. $\dfrac{dQ}{dt}\propto \left( \theta -{{\theta }_{0}} \right)$.
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 $\dfrac{dQ}{dt}=e\sigma \left( {{T}^{4}}-T_{0}^{4} \right)$ ….. (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).
 $\Rightarrow \dfrac{dQ}{dt}=e\sigma \left( {{\left( {{T}_{0}}+\Delta T \right)}^{4}}-T_{0}^{4} \right)$.
$\Rightarrow \dfrac{dQ}{dt}=e\sigma \left( T_{0}^{4}{{\left( 1+\dfrac{\Delta T}{{{T}_{0}}} \right)}^{4}}-T_{0}^{4} \right)$ ….. (iii).
When we have a term ${{\left( 1+x \right)}^{4}}$, x is a very small number (close to zero), the term is approximately equal to $(1+nx)$.
i.e. ${{\left( 1+x \right)}^{n}}\approx 1+nx$.
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).
 $\Rightarrow \dfrac{dQ}{dt}=e\sigma \left( T_{0}^{4}\left( 1+4\dfrac{\Delta T}{{{T}_{0}}} \right)-T_{0}^{4} \right)$
$\Rightarrow \dfrac{dQ}{dt}=e\sigma \left( T_{0}^{4}+4\Delta TT_{0}^{3}-T_{0}^{4} \right)$
$\Rightarrow \dfrac{dQ}{dt}=e\sigma \left( 4\Delta TT_{0}^{3} \right)$
$\Rightarrow \dfrac{dQ}{dt}=e\sigma 4T_{0}^{3}\left( T-{{T}_{0}} \right)$.
Since $e\sigma 4T_{0}^{3}$ is a constant value, $\dfrac{dQ}{dt}\propto \left( T-{{T}_{0}} \right)$.
Hence, proved that the Stefan-Boltzmann’s law is the same as that of Newton’s law of cooling for small temperature difference.

Therefore, the correct option is C.

Note: Let us understand what the other laws given in the options tell us.
(i) Kirchoff’s law: It says that the ratio of emissive power to absorptive power is the same for all surfaces at the same temperature and is equal to the emissive power of a perfectly black body at that temperature.
(ii) Wien’s law: According to this law, the product of wavelength for which a black body emits radiation of maximum intensity and the temperature of the body is constant.