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# A body cools from $50^\circ C$ to $40^\circ C$ in 5 minutes. The surrounding temperature is $20^\circ C$. By how much $^\circ C$ does the temperature decrease in the next 5 minutes? Round your answer to the nearest integer.

Last updated date: 19th Jun 2024
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Hint:The Newton’s law of cooling is defined as the loss of heat of the body is directly proportional to the difference of the temperature of the body and the surrounding. Also the coefficient of heat transfer is constant to the.

Formula used:The formula of the Newton’s law of cooling is given by,
$\Rightarrow \dfrac{{\Delta \theta }}{{\Delta t}} = - k\left( {\bar \theta - {\theta _o}} \right)$
Where $\Delta \theta = {\theta _1} - {\theta _2}$, change in temperature is $\Delta t$, the constant is k, the $\bar \theta$ is the arithmetic mean and ${\theta _o}$ is the surrounding temperature.

Complete step by step solution:
It is given in the problem that a body cools from $50^\circ C$ to $40^\circ C$ in 5 minutes, the surrounding temperature is $20^\circ C$. We need to tell the decrease in temperature in the next 5 minutes.
The formula of the Newton’s law of cooling is given by,

$\Rightarrow \dfrac{{\Delta \theta }}{{\Delta t}} = - k\left( {\bar \theta - {\theta _o}} \right)$

Where $\Delta \theta = {\theta _1} - {\theta _2}$, change in temperature is $\Delta t$, the constant is k, the $\bar \theta$ is the arithmetic mean and ${\theta _o}$ is the surrounding temperature.
Body cools from $50^\circ C$ to $40^\circ C$ in 5 minutes and the surrounding temperature is $20^\circ C$, replacing all these values in Newton's law of cooling.

$\Rightarrow \dfrac{{\Delta \theta }}{{\Delta t}} = - k\left( {\bar \theta - {\theta _o}} \right)$
$\Rightarrow \dfrac{{50 - 40}}{5} = - k\left( {\dfrac{{50 + 40}}{2} - 20} \right)$
$\Rightarrow \dfrac{{10}}{5} = - k\left( {\dfrac{{90}}{2} - 20} \right)$
$\Rightarrow 2 = - k\left( {45 - 20} \right)$
$\Rightarrow 2 = - k\left( {25} \right)$
$\Rightarrow k = \dfrac{{ - 2}}{{25}}$.

Let us calculate the temperature decrease in the next 5 min.
The formula of the Newton’s law of cooling is given by,

$\Rightarrow \dfrac{{\Delta \theta }}{{\Delta t}} = - k\left( {\bar \theta - {\theta _o}} \right)$

Where $\Delta \theta = {\theta _1} - {\theta _2}$, change in temperature is $\Delta t$, the constant is k, the $\bar \theta$ is the arithmetic mean and ${\theta _o}$ is the surrounding temperature.

$\Rightarrow \dfrac{{\Delta \theta }}{{\Delta t}} = - k\left( {\bar \theta - {\theta _o}} \right)$
$\Rightarrow \dfrac{{40 - t}}{5} = \dfrac{2}{{25}}\left( {\dfrac{{40 + t}}{2} - 20} \right)$
$\Rightarrow \dfrac{{40 - t}}{5} = \dfrac{2}{{25}}\left( {\dfrac{{40 + t - 40}}{2}} \right)$
$\Rightarrow 40 - t = \dfrac{{40 + t - 40}}{5}$
$\Rightarrow 5\left( {40 - t} \right) = t$
$\Rightarrow 200 - 5t = t$
$\Rightarrow 6t = 200$
$\Rightarrow t = \dfrac{{200}}{6}$
$\Rightarrow t = 33 \cdot 33^\circ C$.

The final temperature in the next 5 min. is equal $t = 33 \cdot 33^\circ C$.

Note:The students are advised to understand and remember the formula of Newton's law of cooling as it is very helpful in solving problems like these. Newton's law of cooling is the concept which tells us the ratio of heat transfer from one body to another.