Question
A metal sphere cools from $62^{\circ}\text C $ to $50^{\circ}\text C $ in first 10 minutes and in next 10 minutes it cools up to $42^{\circ}\text C.$ Calculate its temperature next 10 minutes according to Newton law of cooling.
✓
Answer
According to Newton law of cooling,
$\frac{T_1-T_2}{t}=K\left[\frac{T_1+T_2}{2}-T_0\right]$
According to question,
$\begin{array}{l}T_1=62^{\circ} C, T_2=50^{\circ} C, t=10 \text { minute } \\\therefore \frac{62-50}{10}=K\left[\frac{62+50}{2}-T_0\right] \\\Rightarrow\left[\frac{12}{10}\right]=K\left[56-T_0\right]\ldots\ldots (1)\end{array}$
In next 10 minutes, $T_1=50^{\circ} C , T_2=42^{\circ} C$
$\begin{array}{l}\therefore \frac{50-42}{10}=K\left[\frac{50+42}{2}-T_0\right] \\\Rightarrow \frac{8}{10}=K\left[46-T_0\right]\ldots\ldots (2)\end{array}$
Let in next $t=10$ minutes if temperature bacome $T$, then
$\frac{42-T}{10}=K\left[\frac{42+T}{2}-T_0\right]\ldots\ldots (3)$
Dividing equation (1) by (2)
$\begin{array}{ll}& \frac{12}{10} \times \frac{10}{8}=\frac{K\left(56-T_0\right)}{K\left(46 - T_0\right)}=\left(\frac{56-T_0}{46-T_0}\right) \\& \frac{3}{2}=\frac{56-T_0}{46-T_0} \\\Rightarrow & 3\left(46-T_0\right)=2\left(56-T_0\right) \\\Rightarrow & 138-3 T_0=112-2 T_0 \\\Rightarrow \quad & 138-112=-2 T_0+3 T_0 \\\therefore \quad & T_0=26\end{array}$
Putting the value of $T_0$ in eq. (1)
$\begin{array}{l}\frac{12}{10}=K(56-26)=30 K \\\therefore K=\frac{12}{10 \times 30}=\frac{12}{300}=\frac{1}{25} \\K=\frac{1}{25}\end{array}$
Putting the values of $T _6$ and K in eq. (3)
$\Rightarrow \quad \frac{42-T}{10}=K\left[\frac{42+T}{2}-T_0\right]$
$\begin{array}{l}\frac{42- T }{10}=\frac{1}{25}\left[\frac{42+ T }{2}-26\right] \\ \frac{42- T }{10}=\frac{1}{25} \times \frac{ T -10}{2}=\frac{ T -10}{50} \\ \Rightarrow \frac{42- T }{10}=\frac{ T -10}{50}\end{array}$
$\Rightarrow \quad \frac{42- T }{1}=\frac{ T -10}{5}$
$\Rightarrow \quad 210-5 T= T -10$
$\begin{array}{ll}\Rightarrow & 220=6 T \\ \Rightarrow & T =\frac{220}{6}=36.36^{\circ} C \end{array}$
$\frac{T_1-T_2}{t}=K\left[\frac{T_1+T_2}{2}-T_0\right]$
According to question,
$\begin{array}{l}T_1=62^{\circ} C, T_2=50^{\circ} C, t=10 \text { minute } \\\therefore \frac{62-50}{10}=K\left[\frac{62+50}{2}-T_0\right] \\\Rightarrow\left[\frac{12}{10}\right]=K\left[56-T_0\right]\ldots\ldots (1)\end{array}$
In next 10 minutes, $T_1=50^{\circ} C , T_2=42^{\circ} C$
$\begin{array}{l}\therefore \frac{50-42}{10}=K\left[\frac{50+42}{2}-T_0\right] \\\Rightarrow \frac{8}{10}=K\left[46-T_0\right]\ldots\ldots (2)\end{array}$
Let in next $t=10$ minutes if temperature bacome $T$, then
$\frac{42-T}{10}=K\left[\frac{42+T}{2}-T_0\right]\ldots\ldots (3)$
Dividing equation (1) by (2)
$\begin{array}{ll}& \frac{12}{10} \times \frac{10}{8}=\frac{K\left(56-T_0\right)}{K\left(46 - T_0\right)}=\left(\frac{56-T_0}{46-T_0}\right) \\& \frac{3}{2}=\frac{56-T_0}{46-T_0} \\\Rightarrow & 3\left(46-T_0\right)=2\left(56-T_0\right) \\\Rightarrow & 138-3 T_0=112-2 T_0 \\\Rightarrow \quad & 138-112=-2 T_0+3 T_0 \\\therefore \quad & T_0=26\end{array}$
Putting the value of $T_0$ in eq. (1)
$\begin{array}{l}\frac{12}{10}=K(56-26)=30 K \\\therefore K=\frac{12}{10 \times 30}=\frac{12}{300}=\frac{1}{25} \\K=\frac{1}{25}\end{array}$
Putting the values of $T _6$ and K in eq. (3)
$\Rightarrow \quad \frac{42-T}{10}=K\left[\frac{42+T}{2}-T_0\right]$
$\begin{array}{l}\frac{42- T }{10}=\frac{1}{25}\left[\frac{42+ T }{2}-26\right] \\ \frac{42- T }{10}=\frac{1}{25} \times \frac{ T -10}{2}=\frac{ T -10}{50} \\ \Rightarrow \frac{42- T }{10}=\frac{ T -10}{50}\end{array}$
$\Rightarrow \quad \frac{42- T }{1}=\frac{ T -10}{5}$
$\Rightarrow \quad 210-5 T= T -10$
$\begin{array}{ll}\Rightarrow & 220=6 T \\ \Rightarrow & T =\frac{220}{6}=36.36^{\circ} C \end{array}$
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