Question

A body takes 5 minutes to cool from $100^{\circ }C$ to $60^{\circ }C$ and takes time $t$ to $\mathrm{cool}$ from 80 to $40^{\circ }C$. If surrounding temperature is $20^{\circ }C$, Find t:

• A. $9min$
• B. $4min$
• C. 6 min
• D. 8 min

Answer 1

Answer: (D)

$t_0=\frac{2.303}{k}{\log }_{10}\frac{T_1-T_0}{T_2-T_0}\left(\rho =\frac{2.303}{k}\right)$
$5=\rho {\log }_{10}\frac{100-20}{60-20}$
$5=\rho {\log }_{10}\frac{80}{40}$
$\Rightarrow 5=\rho {\log }_{10}(2)\dots$(1)
$t=\rho {\log }_{10}\frac{80-20}{40-20}$
$\Rightarrow t=\rho {\log }_{10}\frac{60}{20}$
$t=\rho {\log }_{10}(3)\dots$ (2)
from eq $(1)\&(2)$ we get
$\frac{5}{t}=\frac{A{\log }_{10}(2)}{A{\log }_{10}(3)}$
$t=5\times 1.6=8min$