Question

If an object cools down from 80°C to 60°C in 5 min in a surrounding of temperature 20°C. The time taken to cool from 60°C to 40°C will be (Assume Newton’s law of cooling to be valid)

• A. \(\frac{25}{3}\) min
• B. 5 min
• C. \(\frac{25}{4}\) min
• D. 5 min

Answer 1

The Correct Option is A

Solution:
We will use Newton's Law of Cooling. $$ \frac{dT}{dt} = -k(T - T_s) $$ where:

$T$ is the temperature of the object at time $t$

$T_s$ is the temperature of the surroundings

$k$ is a positive constant

Cooling from $80^{\circ}C$ to $60^{\circ}C$:

Initial temperature $T_1 = 80^{\circ}C$

Final temperature $T_2 = 60^{\circ}C$

Time $t_1 = 5 \text{ minutes} = 300 \text{ seconds}$

Surrounding temperature $T_s = 20^{\circ}C$

Using the average temperature: $$ \frac{T_1 + T_2}{2} = \frac{80 + 60}{2} = 70^{\circ}C $$ Applying Newton's Law: $$ \frac{T_2 - T_1}{t_1} = -k\left(\frac{T_1 + T_2}{2} - T_s\right) $$ $$ \frac{60 - 80}{300} = -k(70 - 20) $$ $$ \frac{-20}{300} = -k(50) $$ $$ k = \frac{20}{300 \cdot 50} = \frac{2}{1500} = \frac{1}{750} $$ Cooling from $60^{\circ}C$ to $40^{\circ}C$:

Initial temperature $T_1 = 60^{\circ}C$

Final temperature $T_2 = 40^{\circ}C$

Time $t_2$ (to be found)

Surrounding temperature $T_s = 20^{\circ}C$

Using the average temperature: $$ \frac{60 + 40}{2} = 50^{\circ}C $$ Applying Newton's Law: $$ \frac{40 - 60}{t_2} = -k(50 - 20) $$ $$ \frac{-20}{t_2} = -k(30) $$ $$ t_2 = \frac{20}{k \cdot 30} = \frac{20}{\frac{1}{750} \cdot 30} = \frac{20 \cdot 750}{30} = 20 \cdot 25 = 500 \text{ seconds} $$ Therefore, the time it takes to cool from $60^{\circ}C$ to $40^{\circ}C$ is 500 seconds. 

The correct answer is (1) 500 s.