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

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) \ldots \ldots( i ) $
$t =\frac{2.303}{ K } \log _{10} \frac{40-20}{30-20}$
$\Rightarrow t =\frac{2.303}{ K } \log _{10} \frac{20}{10} $
$t =\frac{2.303}{ K } \log _{10}(2)\ldots . .(ii)$
From eq (i) $\&$ (ii) we get
$\frac{6}{t}=\frac{\log _{10}(3)}{\log _{10}(2)}$
$t=6 \times 0.631=3.7$ minutes