Question

A pan filled with hot food cools from ${94^ \circ }C$ to ${86^ \circ }C$ in 2 minutes. When the room temperature is ${20^ \circ }C$. How long will it cool from ${74^ \circ }C$ to ${66^ \circ }C$?
A. 2 minutes
B. 2.8 minutes
C. 2.5 minutes
D. 1.8 minutes

Answer 1

Hint: This problem is based on Newton’s Law of Cooling.
Newton’s Law of Cooling states that: the rate of heat loss of a body is directly proportional to the difference in the temperatures between the body and its surroundings.
In mathematical form,
\[
  \dfrac{{\Delta T}}{{\Delta t}} = - K\left( {{T_{avg}} - {T_0}} \right) \\
    \\
 \]
where
\[\dfrac{{\Delta T}}{{\Delta t}}\] is the rate of change of temperature with time
$K$ is a constant
\[{T_{avg}}\& {T_o}\] are the average temperatures and the temperature of surroundings respectively.

Complete step by step solution:
Step 1: Find the constant K
The pan being used to heat is the same. Hence, the constant will be the same. The constant K depends on the material and surface area. Hence, we can find the constant K and substitute to get our answer.
Given data –
$\Delta T = {94^ \circ }C - {86^ \circ }C = {8^ \circ }C$
$\Delta t = 2\min $
\[{T_{avg}} = \left( {\dfrac{{94 + 86}}{2}} \right) = {90^ \circ }C\]
${T_o} = {20^ \circ }C$
Substituting the values in the Newton’s Law of Cooling, we get –
\[
  \dfrac{{\Delta T}}{{\Delta t}} = - K\left( {{T_{avg}} - {T_0}} \right) \\
  \dfrac{8}{2} = - K\left( {90 - 20} \right) \\
  Solving, \\
  4 = - K(70) \\
  K = - \dfrac{4}{{70}} \\
 \]

Step 2: Substitute K for the new condition
Now, we must substitute the value of K in the equation again for the new case –
$\Delta T = {74^ \circ }C - {66^ \circ }C = {8^ \circ }C$
\[{T_{avg}} = \left( {\dfrac{{74 + 66}}{2}} \right) = {70^ \circ }C\]
${T_o} = {20^ \circ }C$
\[K = - \dfrac{4}{{70}}\]
Substituting the values in Newton’s Law of Cooling and solving for $\Delta t$
\[
  \dfrac{{\Delta T}}{{\Delta t}} = - K\left( {{T_{avg}} - {T_0}} \right) \\
  \dfrac{8}{{\Delta t}} = - \left( { - \dfrac{4}{{70}}} \right)\left( {70 - 20} \right) \\
  Solving, \\
  \dfrac{{8 \times 70}}{{4 \times \Delta t}} = 50 \\
  \Delta t = \dfrac{{{8}2 \times 7{0}}}{{{4} \times 5{0}}} = \dfrac{{14}}{5} = 2.8\min \\
 \]

$\therefore$ The time taken for cooling = 2.8 min. Hence, the correct option is Option (B).

Note:
Here, I have directly, substituted the value of K. However, the constant K is the product of the coefficient of heat transfer and surface area.
$
  K = H \times A \\
  where \\
 $
H = heat transfer coefficient of the material.
A = surface area of heat transfer in ${m^2}$