Question

an object is cooled from 75c to 65c in 2min in a room at 30c the time taken to cool the same object from 55c to 45c in the same room is

Answer 1

Given:

• An object is cooled from $75^\circ C$ to $65^\circ C$ in $2min$ in a room at $30^\circ C.$

To Find:

• The time taken to cool the same object from $55^\circ C$ to $45^\circ C$ in the same room

Solution:

Newton’s law of cooling is given by,

  ⇒           $\frac{dT}{dt} =k(T-T_0 )$

$T =$ temperature at time t and

$T_0$ = temperature of the surrounding,

$k$ = Positive constant that depends on the area and nature of the surface of the body under consideration.

Here,

For the interval in which temperature falls from $75^\circ C$ to $65^\circ C$

             $T = \frac{75 + 65}{2}$

             $T = 70^\circ C$

           $T_0 = 30^\circ C$

Putting the values,

             ⇒       $\frac{dT}{dt} = \frac{dT}{2} K(70-30)$

             ⇒      $\frac{dT}{K} =40 \times 2=80$

Now,

    For the interval in which temperature falls from $55^\circ C$ to $45^\circ C$

                              $T = \frac{55 + 45}{2}$

                              $T = 50^\circ C$

        ⇒             $\frac{dT}{dt} =K(50-30)$

            In both cases, $dT$ is same

          ⇒           $dt= \frac{dT}{K} \times \frac{1}{20}$

                                $=\frac{80}{20} =4min$

Thus,  $4min$ is taken to cool the same object from $55^\circ C$ to $45^\circ C$ in the same room.