A body is allowed to cool. It takes $15$ minutes to cool from $80^\circ C$ to $60^\circ C$ The temperature of body in next $15$ minutes would be (Temperature of surrounding is $20^\circ C$ )
A. $41.6^\circ C$
B. $46.7^\circ C$
C. $49.2^\circ C$
D. $52^\circ C$
Answer
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Hint: In order to solve this question we need to understand Newton’s law of cooling which states that rate of heat loss of a body is directly proportional to difference in temperature between bodies and surrounding. Here Newton considered only one method of heat flow and that is radiation so he completely neglected other forms of heat transfer known as convection and conduction. Conduction involves heat transfer through phonons of vibration but radiation involves photons of light to convey the heat.
Complete step by step answer:
Given, Temperature of surrounding, ${T_0} = 20^\circ C$. Let the body initially take ${t_1} = 15\min $ to cool from temperature, ${T_1} = 80^\circ C$ to temperature, ${T_2} = 60^\circ C$. Using the formula of Newton’s cooling law,
$t = \dfrac{1}{k}\log (\dfrac{{{T_1} - {T_0}}}{{{T_2} - {T_0}}})$
Putting values we get,
$15 = \dfrac{1}{k}\log (\dfrac{{80 - 20}}{{60 - 20}})$
$\Rightarrow 15 = \dfrac{1}{k}\log (\dfrac{3}{2})$
$\Rightarrow k = \dfrac{1}{{15}} \times 0.176$
$\Rightarrow k = 0.0117\,{\min ^{ - 1}}$
Here, $k$ is proportionality constant.
Next time, ${t_2} = 15\min $ the body cools from ${T_1}' = 60^\circ C$ to temperature, ${T_2}' = T^\circ C$. Putting values we get,
${t_2} = \dfrac{1}{k}\log (\dfrac{{{T_1}' - {T_0}}}{{{T_2}' - {T_0}}}) \\ $
$\Rightarrow 15 = \dfrac{1}{{0.0177}}\log (\dfrac{{60 - 20}}{{T - 20}}) \\ $
$\Rightarrow \log (\dfrac{{40}}{{T - 20}}) = 0.0117 \times 15 \\$
$\Rightarrow \log (\dfrac{{40}}{{T - 20}}) = 0.1755 \\ $
$\Rightarrow \dfrac{{40}}{{T - 20}} = 1.5$
Cross multiplication we get,
$40 = 1.5T - 30 \\ $
$\Rightarrow T = \dfrac{{70}}{{1.5}} \\ $
$\therefore T = 46.667$
So the correct option is B.
Note: It should be remembered that Heat is the thermal energy which only flows due to temperature difference between system and surrounding. There are three methods of heat flow, one is conduction in which heat flows due to atoms vibration and exchange of heat occurs through phonons of vibration, another is convection in which heat transports through matter exchange like convection current flows in summer, and third is radiation in which heat flows due to exchange of photons in this heat flow like in blackbody radiation.
Complete step by step answer:
Given, Temperature of surrounding, ${T_0} = 20^\circ C$. Let the body initially take ${t_1} = 15\min $ to cool from temperature, ${T_1} = 80^\circ C$ to temperature, ${T_2} = 60^\circ C$. Using the formula of Newton’s cooling law,
$t = \dfrac{1}{k}\log (\dfrac{{{T_1} - {T_0}}}{{{T_2} - {T_0}}})$
Putting values we get,
$15 = \dfrac{1}{k}\log (\dfrac{{80 - 20}}{{60 - 20}})$
$\Rightarrow 15 = \dfrac{1}{k}\log (\dfrac{3}{2})$
$\Rightarrow k = \dfrac{1}{{15}} \times 0.176$
$\Rightarrow k = 0.0117\,{\min ^{ - 1}}$
Here, $k$ is proportionality constant.
Next time, ${t_2} = 15\min $ the body cools from ${T_1}' = 60^\circ C$ to temperature, ${T_2}' = T^\circ C$. Putting values we get,
${t_2} = \dfrac{1}{k}\log (\dfrac{{{T_1}' - {T_0}}}{{{T_2}' - {T_0}}}) \\ $
$\Rightarrow 15 = \dfrac{1}{{0.0177}}\log (\dfrac{{60 - 20}}{{T - 20}}) \\ $
$\Rightarrow \log (\dfrac{{40}}{{T - 20}}) = 0.0117 \times 15 \\$
$\Rightarrow \log (\dfrac{{40}}{{T - 20}}) = 0.1755 \\ $
$\Rightarrow \dfrac{{40}}{{T - 20}} = 1.5$
Cross multiplication we get,
$40 = 1.5T - 30 \\ $
$\Rightarrow T = \dfrac{{70}}{{1.5}} \\ $
$\therefore T = 46.667$
So the correct option is B.
Note: It should be remembered that Heat is the thermal energy which only flows due to temperature difference between system and surrounding. There are three methods of heat flow, one is conduction in which heat flows due to atoms vibration and exchange of heat occurs through phonons of vibration, another is convection in which heat transports through matter exchange like convection current flows in summer, and third is radiation in which heat flows due to exchange of photons in this heat flow like in blackbody radiation.
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