A pan filled with hot food cools from 50-1 - C to 49-9 - C in 5 sec- How long in seconds will it take to cool from 40-1 - C to 39-9 - C if room temperature is 30 - C-

The mathematical interpretation of Newton's law of cooling states that: dTdt = bA(T T_0) ; where T_0 is room temperature Solving this differential equ ...

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Byju's Answer

Standard XII

Physics

Time Period Independent of Amplitude

A pan filled ...

Question

A pan filled with hot food cools from ${50.1}^{\circ} C$ to ${49.9}^{\circ} C$ in $5 s e c$ . How long in seconds will it take to cool from ${40.1}^{\circ} C$ to ${39.9}^{\circ} C$ if room temperature is $30^{\circ} C$ ?

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Solution

The mathematical interpretation of Newton's law of cooling states that:
$\frac{d T}{d t} = - b A (T - T_{0})$ ; where $T_{0} i s r o o m t e m p e r a t u r e$
Solving this differential equation, with basic integration and applying the limits:
$T$ ; varying from $T_{1} t o T_{2}$
$t$ ; varying from 0 to t. The time it takes to change from $T_{1} t o T_{2}$
Thus we get:
$\frac{1}{t} l n (\frac{T_{2} - T_{0}}{T_{1} - T_{0}}) = - b A = c o n s t a n t$
Here, 't' is the time taken to for cooling from $T_{1}$ to $T_{2}$ , $T_{0}$ is the room temperature.
Now applying this here:
$\frac{1}{5} l n (\frac{19.9}{20.1}) = \frac{1}{t} l n (\frac{9.9}{10.1})$
Solving this we get t = 10s

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A pan filled with hot food cools from 50.1∘C to 49.9∘C in 5sec. How long in seconds will it take to cool from 40.1∘C to 39.9∘C if room temperature is 30∘C?

A pan filled with hot food cools from ${50.1}^{\circ} C$ to ${49.9}^{\circ} C$ in $5 s e c$ . How long in seconds will it take to cool from ${40.1}^{\circ} C$ to ${39.9}^{\circ} C$ if room temperature is $30^{\circ} C$ ?