Đáp án:

${t_{cb}}'' = {35^o}C$ 

Giải thích các bước giải:

Ta có:

$\begin{array}{l}
{Q_1} = {Q_2}\\
 \Leftrightarrow {m_1}{c_1}\left( {{t_{cb}} - {t_1}} \right) = {m_2}{c_2}\left( {{t_2} - {t_{cb}}} \right)\\
 \Leftrightarrow \dfrac{{{m_2}{c_2}}}{{{m_1}{c_1}}} = \dfrac{{30 - 20}}{{{t_2} - 30}} = \dfrac{{10}}{{{t_2} - 30}}\left( 1 \right)
\end{array}$

Ở lần thêm thứ 2:

$\begin{array}{l}
{Q_1}' = {Q_2}'\\
 \Leftrightarrow 2{m_1}{c_1}\left( {{t_{cb}}' - {t_1}} \right) = {m_2}{c_2}\left( {{t_2} - {t_{cb}}'} \right)\\
 \Leftrightarrow \dfrac{{{m_2}{c_2}}}{{{m_1}{c_1}}} = \dfrac{{2.\left( {26 - 20} \right)}}{{{t_2} - 26}} = \dfrac{{12}}{{{t_2} - 26}}\left( 2 \right)\\
\left( 1 \right)\left( 2 \right) \Rightarrow {t_2} = {50^o}C\\
\left( 1 \right) \Rightarrow \dfrac{{{m_2}{c_2}}}{{{m_1}{c_1}}} = \dfrac{{10}}{{50 - 30}} = \dfrac{1}{2}
\end{array}$

Ở lần thêm cuối:

$\begin{array}{l}
{Q_1}'' = {Q_2}''\\
 \Leftrightarrow 2{m_1}{c_1}\left( {{t_{cb}}'' - {t_1}} \right) = 4{m_2}{c_2}\left( {{t_2} - {t_{cb}}''} \right)\\
 \Leftrightarrow \dfrac{{{t_{cb}}'' - {t_1}}}{{{t_2} - {t_{cb}}''}} = \dfrac{{2{m_2}{c_2}}}{{{m_1}{c_1}}} = 2.\dfrac{1}{2} = 1\\
 \Leftrightarrow {t_{cb}}'' - 20 = 50 - {t_{cb}}''\\
 \Leftrightarrow {t_{cb}}'' = {35^o}C
\end{array}$