Question

A body takes 10 minutes to cool from 60°C to 50°C. The temperature of the surroundings is constant at 25°C. Then, the temperature of the body after the next 10 minutes will be approximately
A. \[43^\circ C\]
B. \[47^\circ C\]
C. \[41^\circ C\]
D. \[45^\circ C\]

Answer 1

Hint: Solution to this question will be determined by using Newton’s Law of Cooling. From the given values, we will determine the heat transfer coefficient. The next and final step will be to use the coefficient and the penultimate temperature change to find out the ultimate temperature of the given body.

Formula Used:
Newton’s Law of Cooling: $\dfrac{{d\theta }}{{dt}} = k(\Delta T) = k(\dfrac{{\sum \theta }}{2} - {\theta _0})$
Where $\theta $ is the temperature of the body during instances of change and is expressed in Celsius $(^\circ C)$. It will be determined as \[{\theta _1},{\theta _2},{\theta _3},{\theta _0}\] for the first, second, third and surrounding temperatures respectively. $t$ is the time period for the above temperature changes and ${t_1},{t_2}$ are the time periods for the first and second temperature changes. $dt$ is the temperature change of the body and is expressed in Celsius $(^\circ C)$, $dt$ is the time taken for those changes to occur and is expressed in minutes $(\min )$ and $k$ is heat transfer coefficient.

Complete step by step answer:
We have the following given data:
\[{\theta _1} = 60^\circ C\]
\[{\theta _2} = 50^\circ C\]
\[{\theta _0} = 25^\circ C\]
${t_1} = 10\min $
${t_2} = 10\min $
As we can see, we need the value of $k$ to determine the final temperature change. Therefore using the given data in Newton’s Law of Cooling we get,
$
  \dfrac{{d\theta }}{{dt}} = k(\Delta T) = k(\dfrac{{\sum \theta }}{2} - {\theta _0}) \\
   \Rightarrow \dfrac{{d\theta }}{{dt}} = k(\dfrac{{\sum \theta }}{2} - {\theta _0}) = k(\dfrac{{60 + 50}}{2} - 26) \\
   \Rightarrow \dfrac{{60 - 50}}{{10}} = k(\dfrac{{60 + 50}}{2} - 26) \\
   \Rightarrow k = \dfrac{1}{{29}} \\
 $
Now that we have the value of $k$, we can easily determine \[{\theta _3}\].
Again using Newton’s Law of Cooling and substituting the available data we get,
$
  \dfrac{{d\theta }}{{dt}} = k(\Delta T) = k(\dfrac{{\sum \theta }}{2} - {\theta _0}) \\
   \Rightarrow \dfrac{{50 - {\theta _3}}}{{10}} = \dfrac{1}{{29}}(\dfrac{{50 + {\theta _3}}}{2} - 26) = 44.4117^\circ C \\
   \Rightarrow {\theta _3} \approx 45^\circ C \\
 $

In conclusion, the correct option is D.

Note:the temperature should not be approximated by stepping down. If its $44.4117^\circ C$, it is to be either stepped up or rounded off to $44^\circ C$. Never should it be considered by stepping down to $43^\circ C$.