Question

A hot body placed in a surrounding of temperature ${T_0}$.its temperature at $t = 0$ in ${T_1}$.the specific heat capacity of the body is $s$and its mass is $m$.Assume Newton's law of cooling to be valid. Find:
$(A)$ The measurement head that the body can lose and
$(B)$ The time starting from $t = 0$ in which it will lose $50\%$ of this maximum heat.

Answer 1

We understand the maximum temperature and temperature rate here, so we measure the cooling rate by dividing each temperature data point by its corresponding time data point, then average all the responses to achieve a cooling rate.

Formula used:
This expression represents Newton’s law of cooling. It can be derived directly from Stefan’s law,
$\dfrac{{d\theta }}{{dt}} = {\text{ }} - k[q{\text{ }}-{\text{ }}{{\text{q}}_s}]$
Where,
$q$and ${q_s}$ are temperatures corresponding to objects and surroundings.
The heat is transferred more rapidly as the body temperature changes. Newton’
${q_f}\; = {\text{ }}{q_0}\; + {\text{ }}({q_i}\;-{\text{ }}{q_0}){\text{ }}{e^{\; - k}}$
${q_i}\; =$initial temperature of object
${q_f}\; =$ final temperature of object

Complete step by step solution:
Given by,
Let as take ${T_0} = {\theta _0}$ and ${T_1} = {\theta _1}$
the specific heat capacity of the body is $s$and its mass is $m$
Assume Newton's law of cooling to be valid.
Now, we find the measurement head that the body can lose
According to the Newton cooling Rule,
$\Rightarrow$ $\dfrac{{d\theta }}{{dt}} = {\text{ }} - k[\theta {\text{ }}-{\text{ }}{\theta _0}]$
$(A)$ Highest heat the body will loss,
$\Rightarrow$ $\Delta {Q_{\max }} = ms\left( {{\theta _1} - {\theta _0}} \right)$
$(B)$ If the body loses $50\%$ of maximum heat, the temperature drop will be somewhere.
$\Rightarrow$ $\Delta {Q_{\max }} \times \dfrac{{50}}{{100}} = ms\left( {{\theta _1} - \theta } \right)$
Rearranging the given equation,
We get,
$\Rightarrow$ $ms\left( {{\theta _1} - {\theta _0}} \right) \times \dfrac{1}{2} = ms\left( {{\theta _1} - \theta } \right)$
Cancel the common factor,
$\Rightarrow$ $\theta = {\theta _1} - \left( {{\theta _1} - {\theta _0}} \right) \times \dfrac{1}{2}$
$\Rightarrow$ $\theta = \dfrac{{\left( {{\theta _1} - {\theta _0}} \right)}}{2}$………$(1)$
From Newton’s cooling theory,
$\dfrac{{d\theta }}{{dt}} = {\text{ }} - k[{\theta _1}-\theta ]$
If we integrate this equation within the appropriate limit, we get
At time $t = 0$
$\theta = {\theta _1}$
At time $t$,
$\theta = \theta$
According to the Stefan law,
$\Rightarrow$ $\int {_{{\theta _1}}^\theta } \dfrac{{d\theta }}{{{\theta _1} - \theta }} = - k\int {_0^t} dt$
In $\left( {\dfrac{{{\theta _1} - \theta }}{{{\theta _1} - {\theta _0}}}} \right) = - kt$
We rearranging the above equation
Can be written as,
$\Rightarrow$ ${\theta _1} - \theta = \left( {{\theta _1} - {\theta _0}} \right){e^{ - kt}}$ ………………$(2)$
From equation $(1)$ and $(2)$
$\Rightarrow$ $\dfrac{{{\theta _1} - {\theta _0}}}{2} - {\theta _0} = \left( {{\theta _1} - {\theta _0}} \right){e^{ - kt}}$
$\Rightarrow$ $t = \,In\,\left( {\dfrac{2}{k}} \right)$
The time starting from $t = 0$ in which it will lose $50\%$ of this maximum heat is $t = \,In\,\left( {\dfrac{2}{k}} \right)$.

Note: Hence we can assume a constant rate of cooling, which is equal to the rate of cooling corresponding to the average body temperature during the interval when we only need estimated values from Newton's law. The temperature of the object and its environment, as long as the difference is minimal.