Question

A metallic sphere cools from ${50}^{o}C$ to ${40}^{o}C$ in $300$ seconds. If the room temperature is ${20}^{o}C$ then its temperature in the next $5$ minutes will be:

• A. ${30}^{o}C$
• B. ${33.3}^{o}C$
• C. ${36}^{o}C$
• D. ${38}^{o}C$

Answer 1

The correct option is B ${33.3}^{o}C$

The mathematical interpretation of Newton's law of cooling states that:

$\frac{dT}{dt}=-bA(T-T_0)$; where $T_0$ is room temperature

Solving this differential equation, with basic integration and applying the limits:

$T$; varying from $T_{1}to{T}_2$

$t$; varying from 0 to t. The time it takes to change from $T_{1}to{T}_2$

Thus we get:

$\frac{1}{t}ln\left(\frac{T_2-T_0}{T_1-T_0}\right)=-bA=constant$

Here, 't' is the time taken to for cooling from $T_1$ to $T_2$, $T_0$ is the room temperature.

Applying this here:

$\frac{1}{300}ln\left(\frac{20}{30}\right)=\frac{1}{300}ln\left(\frac{T_2-20}{20}\right)$

i.e.

$ln\left(\frac{2}{3}\right)=ln\left(\frac{T_2-20}{20}\right)$

$\frac{2}{3}=\frac{T_2-20}{20}$

or $T_2={33.33}^{\circ }C$