Question

A pan filled with hot food colds from 94°C to 86°C in 2 minutes when the room temperature is at 20°C. The time taken to cool it from 71°C to 69°C is

Answer 1

Here is the answer of the question.

Answer 2

Answer:

The time taken to cool the pan t= 41.4 sec.

Given:

The pan cools down from 94°C to 86°C in 2 minutes and the room temperature is 20°C.

Writing it in simple format,

T1=94°c

T2=86°c  

T3=20°c  

t’=2 minutes (from 94°C to 86°C)

t=?(from 71°C to 69°C)

Solution:

$\begin{array}{l}{\ln \left(\frac{T 2-T 3}{T 1-T 3}\right)=-k t} \\ {\ln \left(\frac{86-20}{94-20}\right)=-2 k} \\ {\ln \left(\frac{66}{74}\right)=-2 k} \\ {\ln \left(\frac{33}{74}\right) \times\left(-\frac{1}{2}\right)=k} \\ {-\frac{1}{2} \ln \left(\frac{33}{31}\right)=k}\end{array}$

Now to find ‘t’

$\begin{array}{l}{\ln \left(\frac{69-20}{71-20}\right)=-\frac{1}{2} \ln \left(\frac{33}{31}\right) \mathrm{t}} \\ {\ln \left(\frac{49}{51}\right)=-\frac{1}{2} \ln \left(\frac{33}{31}\right) \mathrm{t}} \\ {t=\frac{\ln \left(\frac{49}{51}\right)}{-\frac{1}{2} \ln \left(\frac{3 \mathrm{s}}{31}\right)}}\end{array}$

Solving the above equation, we get,

t= 0.69 minutes  

t=  0.69 × 60  

t=41.4 seconds.