Answer
384.9k+ views
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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)