Đáp án:

$\rm \kern,9cm\gamma ^{\dfrac{T_1-T_2}{10}}=\dfrac{v_1}{v_2}=\dfrac{t_2}{t_1}\\\Leftrightarrow \kern,3cm\gamma^{\dfrac{20-40}{10}}=\dfrac{3}{27}\\\Leftrightarrow \kern,3cm\gamma^{-2}=\dfrac{1}{9}\\\Leftrightarrow \kern,3cm\gamma=3$

Thời gian hoà tan mẫu aluminum trong $\rm 60^°C$:

$\rm \kern,9cm\gamma ^{\dfrac{T_1-T_3}{10}}=\dfrac{v_1}{v_3}=\dfrac{t_3}{t_1}\\\Leftrightarrow \kern,3cm{3}^{\dfrac{20-60}{10}}=\dfrac{t_3}{27}\\\Leftrightarrow \kern,3cmt_3=\dfrac{1}{3}\ (min)$

$\dfrac{v_1}{v_2}$~$\dfrac{t_2}{t_1}=\gamma^{\dfrac{T_1-T_2}{10}}$

`<=>`$\dfrac{3}{27}=\gamma^{\dfrac{20-40}{10}}$

`=>`$\gamma=3$

Tương tự ta có, $\dfrac{t_3}{t_1}=\gamma^{\dfrac{T_1-T_3}{10}}$

`<=>`$\dfrac{t}{27}=3^{\dfrac{20-60}{10}}$

`=>`$t=\dfrac{1}{3}(min)=20(sec)$