Question

A metallic sphere cools from $50{}^{\circ }\text{C}$ to $40{}^{\circ }\text{C}$ in $300\text{s}$. If atmospheric temperature around is $20{}^{\circ }\text{C}$, then the sphere's temperature after the next $5\text{minutes}$ will be close to:

• A. $31{}^{\circ }\text{C}$
• B. $33{}^{\circ }\text{C}$
• C. $28{}^{\circ }\text{C}$
• D. $35{}^{\circ }\text{C}$

Answer 1

The correct option is B $33{}^{\circ }\text{C}$

From Newton's Law of cooling,

$\frac{T_1-T_2}{t}=K[\frac{T_1+T_2}{2}-T_0]$

For case $1:$

Here, $T_1=50{}^{\circ }\text{C},T_2=40{}^{\circ }\text{C}$
and $T_0=20{}^{\circ }\text{C},t=300\text{s}=5\text{minutes}$

$\Rightarrow \frac{50-40}{5\text{min}}=K(\frac{50+40}{2}-20)--\left(1\right)$

Let $T$ be the temperature of the sphere after the next $5\text{minutes}$. Then

$\frac{40-T}{5}=K(\frac{40+T}{2}-20)--\left(2\right)$

Dividing equation $\left(2\right)$ by $\left(1\right)$, we get

$\frac{40-T}{10}=\frac{40+T-40}{50+40-40}=\frac{T}{50}$

$\Rightarrow 40-T=\frac{T}{5}\Rightarrow 200-5T=T$

$\therefore T=\frac{200}{6}={33.3}^{\circ }\text{C}$

Hence, option $\left(B\right)$ is correct.