A pan filled with hot food cools from 94^ C to 86^ C is 2 minutes… - Vidyadip

Question

A pan filled with hot food cools from $94^{\circ}\text C $ to $86^{\circ}\text C $ is 2 minutes when the room temperature is at $20^{\circ}\text C $. How long will it take to cool from $71^{\circ}\text C $ to $69^{\circ}\text C $?

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Answer

The average temperature of $94^{\circ} C$ and $86^{\circ} C$ is $90^{\circ} C$ which is $70^{\circ} C$ more than room temperature. In these conditions the temperature of the vessel decreases in 2 minutes ( $94^{\circ} C -86^{\circ} C =8^{\circ} C$ ).
From Newton's law of cooling we know that
$\begin{array}{c}\frac{\text { Temperature difference }}{\text { Time }}=K \Delta T \\\frac{8^{\circ} C}{2 \text { minute }}=K\left(70^{\circ} C\right)\ldots\ldots (1)\end{array}$
The mean temperature of $69^{\circ} C$ and $71^{\circ} C$ is $70^{\circ} C$ which is $50^{\circ} C$ more than room temperature. In this state K is same as the ground state.
Hence,
$\frac{2^{\circ} C}{\text { Time }}=K\left(50^{\circ} C\right)\ldots\ldots (2)$
Dividing both equation
$\begin{aligned} \frac{8^{\circ} C / 2 \text { minute }}{2^{\circ} C / \text { time }} & =\frac{70^{\circ} C }{50^{\circ} C }=\frac{7}{5} \\ & =\frac{8 \times \text { time }}{2 \times 60 \times 2}=\frac{7}{5}\end{aligned}$
$\begin{array}{ll}\Rightarrow & \text { Time }=\frac{7 \times 2 \times 60 \times 2}{5 \times 8}=42 \text { second } \\ \Rightarrow & \text { Time }=42 \text { second }\end{array}$

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