Question

Newton’s law of cooling is a special case of

• A. Kirchhoff’s law
• B. Wien's law
• C. Stefan-Boltzmann’s law
• D. Planck’s law

Answer 1

The correct option is C

Stefan-Boltzmann’s law

Explanation for the correct option

In case of option (c)

Newton's law of cooling

• Newton's law of cooling states that the rate at which a body loses heat is directly proportional to the temperature difference between the body and its surroundings, provided the temperature difference is very small.
• Mathematically, this law states as follows:

$\frac{-dQ}{dt}\propto (\theta -{\theta }_o)$

Here, $\frac{-dQ}{dt}$ is the rate at which a body loses heat, $\theta$ is the temperature of the cooling body and ${\theta }_o$ is the temperature of the surroundings.

Stefan-Boltzmann's law

• Stefan-Boltzmann's law states that the rate at which a body cools, $\frac{-dQ}{dt}$ is given by

$\frac{-dQ}{dt}=e\sigma \left(T^4-{T_o}^4\right).......\left(1\right)$

Here, $e$ is the emissivity of the body, $\sigma$ is Stefan-Boltzmann's constant, $T$ is the temperature of the cooling body and $T_o$ is the temperature of the surroundings.

• Newton's law of cooling is a special case of Stefan-Boltzmann's law provided the temperature of the body and the surroundings is very small.

Proof of the above statement

Let the temperature difference of the body and the surroundings is $∆T=T-T_o$

$\Rightarrow T=T_{\circ }+∆T$

Substitute the value of $T$ in equation $\left(1\right)$

$$\begin{gathered} \frac{-dQ}{dt}=e\sigma \left({\left(T_o+∆T\right)}^4-{T_o}^4\right) \\ \Rightarrow \frac{-dQ}{dt}=e\sigma \left({T_o}^4{\left(1+\frac{∆T}{T_o}\right)}^4-{T_o}^4\right) \\ \Rightarrow \frac{-dQ}{dt}=e\sigma \left({T_o}^4\left(1+4\frac{∆T}{T_o}\right)-{T_o}^4\right) \\ \Rightarrow \frac{-dQ}{dt}=e\sigma \left({T_o}^4+4{T_o}^3∆T-{T_o}^4\right) \\ \Rightarrow \frac{-dQ}{dt}=e\sigma \left(4∆T{T_o}^3\right) \\ \Rightarrow \frac{-dQ}{dt}=4e\sigma {T_o}^3(T-T_o) \end{gathered}$$ (since $∆T$ is very small, the ratio $\frac{∆T}{T_o}$ is also very small. So, ${\left(1+\frac{∆T}{T_o}\right)}^4\approx 1+4\frac{∆T}{T_o}$)

Since $4e\sigma {T_o}^3$ is a constant quantity, $\frac{-dQ}{dt}\propto \left(T-T_o\right)$, which is Newton's law of cooling.

Explanation for the incorrect options

In case of option (a)

• Kirchhoff's law states that the ratio of emissive power to absorptive power is the same for all surfaces at a constant temperature and is equal to the emissive power of a blackbody at that temperature.
• Therefore, option (a) is incorrect.

In case of option (b)

• Wien's law states that the product of wavelength at which the intensity of the radiation emitted by the blackbody is maximum and the temperature of the body is constant.
• Therefore, option (b) is incorrect.

In case of option (d)

• Planck's law gives the spectral density of electromagnetic radiation.
• These radiation are emitted by a blackbody when they remain in thermal equilibrium at a particular temperature.
• Therefore, option (d) is incorrect.

Hence, option (c) is the correct answer.