Question

A body cools down from 50°C to 45°C in 5 mintues and to 40°C in another 8 minutes. Find the temperature of the surrounding.

Answer 1

The temperature of the surrounding is $T_{o}=34^{\circ} C$

Explanation:

Step 1:

Let the temperature of the ambient be $\mathrm{T}_{0}^{\circ} \mathrm{C}$

First case

Initial body temperature = 50°C

Final body temperature = 45°C

Average body  temperature = $\frac{50+45}{2}=\frac{95}{2}=47.5^{\circ} \mathrm{C}$

Difference between body temperature and surroundings =$(47.5-T)^{\circ} C$

Temperature Fall Rate =

$\frac{\Delta T}{t}=\frac{5}{5}=1^{\circ} \mathrm{C} / \mathrm{min}$

Step 2:

By the Newton Cooling Principle or law ,

$\frac{d T}{d t}=-\mathrm{K}\left[T_{\mathrm{avg}}-T_{0}\right]$

$1=-\mathrm{k}\left[47.5-\mathrm{T}_{0}\right]$       ……..  eqn (1)

Second case

Initial body  temperature = 45°C

Final body  temperature = 40°C

Average  body temperature = $\frac{50+45}{2}=\frac{85}{2}=42.5^{\circ} \mathrm{C}$

Difference between body temperature and surroundings = $\left(42.5-T_{0}\right)^{\circ} \mathrm{C}$

Temperature Fall Rate =

$\frac{\Delta T}{t}=\frac{5}{8}=\frac{5}{8} \frac{c}{m i n}$

Step 3:

By the Newton Cooling Principle or law ,

$\frac{d T}{d t}=-K\left[T_{a v g}-T_{o}\right]$

$0.625=-K\left[42.5-T_{o}\right]$      ….…… eqn (2)

Dividing equation  (1) by equation (2),

$\frac{1}{0,625}=\frac{\left[47.5-T_{0}\right]}{\left[42.5-T_{0}\right]}$

$42.5-T_{o}=29.68-0.625 T_{o}$

$42.5-29.68=-0.625 T_{o}+T_{o}$

$0.375 T_{o}=12.82$

$T_{o}=\frac{12.82}{0.375}$

$T_{o}=34^{\circ} C$