
If a body coated black at $ 600\,k $ surrounded by atmosphere at, $ 300\,k $ has cooling rate $ {r_1} $ the same body at $ 900\,k $ surrounded by the same atmosphere, will have cooling rate equal to
$A.\dfrac{{16}}{3}\,{r_1} $
$B.\dfrac{8}{{16}}\,{r_0}$
$C.16\,{r_0}$
$D.4\,{r_0}$
Answer
579.6k+ views
Hint
In the question the atmosphere temperature and body temperature is given. Substitute the values in the equation of Newton's law of cooling, we get the value of cooling rate of the temperature.
The expression for finding the rate of cooling is
$ r \propto \,{T^4} - {T_0}^4 $
Where, $ r $ be the rate of the cooling, $ T $ be the body temperature, $ {T_0} $ be the temperature of the atmosphere.
Complete step by step answer
In the question we know that the same atmosphere temperature is given to the two body temperatures. So we written the expression as
$ \dfrac{{{r_2}}}{{{r_1}}} = \left( {\dfrac{{{T_2}^4 - {T_0}^4}}{{{T_1}^4 - {T_0}^4}}} \right).........\left( 1 \right) $
Where, $ {T_1} $ be the first body temperature, $ {T_2} $ be the second body temperature.
Given that $ {T_1} = 600\,K,{T_2} = 900\,K,{T_0} = 300\,K $
Substitute the known values in the equation $ \left( 1 \right) $ ,we get
$ \dfrac{{{r_2}}}{{{r_1}}} = \left( {\dfrac{{{{900}^4} - {{300}^4}}}{{{{600}^4} - {{300}^4}}}} \right) $
Solving the above equation, we get
$ \dfrac{{{r_2}}}{{{r_1}}} = \left( {\dfrac{{{9^4} - {3^4}}}{{{6^4} - {3^4}}}} \right) $
Simplify the above equation, we get
$ \dfrac{{{r_2}}}{{{r_1}}} = \left( {\dfrac{{{{81}^2} - {9^2}}}{{{{36}^2} - {9^2}}}} \right) $
$ \dfrac{{{r_2}}}{{{r_1}}} = \left( {\dfrac{{16}}{3}} \right) $
Solving the above equation we get $ {r_2} $ as,
$ {r_2} = \dfrac{{16}}{3}{r_1} $
Therefore, from the above option, option (A) is correct.
Note
From the equation of Newton's law of cooling, the rate of cooling is proportional to the temperature of the body and the surroundings. But in the question, there are two rates of cooling that are given so that we divide the highest by the lowest. We want to find the ratio of cooling for the second temperature so we convert the rate of cooling in terms of the second temperature. Here the temperature is measured in Kelvin
In the question the atmosphere temperature and body temperature is given. Substitute the values in the equation of Newton's law of cooling, we get the value of cooling rate of the temperature.
The expression for finding the rate of cooling is
$ r \propto \,{T^4} - {T_0}^4 $
Where, $ r $ be the rate of the cooling, $ T $ be the body temperature, $ {T_0} $ be the temperature of the atmosphere.
Complete step by step answer
In the question we know that the same atmosphere temperature is given to the two body temperatures. So we written the expression as
$ \dfrac{{{r_2}}}{{{r_1}}} = \left( {\dfrac{{{T_2}^4 - {T_0}^4}}{{{T_1}^4 - {T_0}^4}}} \right).........\left( 1 \right) $
Where, $ {T_1} $ be the first body temperature, $ {T_2} $ be the second body temperature.
Given that $ {T_1} = 600\,K,{T_2} = 900\,K,{T_0} = 300\,K $
Substitute the known values in the equation $ \left( 1 \right) $ ,we get
$ \dfrac{{{r_2}}}{{{r_1}}} = \left( {\dfrac{{{{900}^4} - {{300}^4}}}{{{{600}^4} - {{300}^4}}}} \right) $
Solving the above equation, we get
$ \dfrac{{{r_2}}}{{{r_1}}} = \left( {\dfrac{{{9^4} - {3^4}}}{{{6^4} - {3^4}}}} \right) $
Simplify the above equation, we get
$ \dfrac{{{r_2}}}{{{r_1}}} = \left( {\dfrac{{{{81}^2} - {9^2}}}{{{{36}^2} - {9^2}}}} \right) $
$ \dfrac{{{r_2}}}{{{r_1}}} = \left( {\dfrac{{16}}{3}} \right) $
Solving the above equation we get $ {r_2} $ as,
$ {r_2} = \dfrac{{16}}{3}{r_1} $
Therefore, from the above option, option (A) is correct.
Note
From the equation of Newton's law of cooling, the rate of cooling is proportional to the temperature of the body and the surroundings. But in the question, there are two rates of cooling that are given so that we divide the highest by the lowest. We want to find the ratio of cooling for the second temperature so we convert the rate of cooling in terms of the second temperature. Here the temperature is measured in Kelvin
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