Question

A beaker of boiled water cools from $80^{\circ }C$ to $40^{\circ }C$ in 6 minutes. What is the time taken to cool from $40^{\circ }C$ to $30^{\circ }C.$ The temperature of the surrounding is $20^{\circ }c$

• A. $8min$
• B. $3.7min$
• C. $5min$
• D. $4min$

Answer 1

Answer: (B)

$t_0=\frac{2.3026}{K}{\log }_{10}\frac{T_1-T_0}{T_2-T_0}$
$t_0=\frac{2.3026}{K}{\log }_{10}\frac{80-20}{40-20}$
$\Rightarrow t_0=\frac{2.3026}{K}{\log }_{10}\frac{60}{20}$
$6=\frac{2.303}{K}{\log }_{10}(3)\dots \dots (i)$
$t=\frac{2.303}{K}{\log }_{10}\frac{40-20}{30-20}$
$\Rightarrow t=\frac{2.303}{K}{\log }_{10}\frac{20}{10}$
$t=\frac{2.303}{K}{\log }_{10}(2)\dots ..(ii)$
From eq (i) $\&$ (ii) we get
$\frac{6}{t}=\frac{{\log }_{10}(3)}{{\log }_{10}(2)}$
$t=6\times 0.631=3.7$ minutes