Question

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

• A. $9 min$
• B. $4 min$
• 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
$\frac{5}{t}=\frac{A \log _{10}(2)}{A \log _{10}(3)}$
$t=5 \times 1.6=8 min$