Đáp án:

$t' = 73,{33^o}C$

Giải thích các bước giải:

 Nhiệt độ cân bằng là:
$\begin{array}{l}
{Q_{thu}} = {Q_{toa}}\\
 \Leftrightarrow {m_1}{c_1}\left( {t - {t_1}} \right) + mc\left( {t - {t_1}} \right) = 2mc\left( {{t_3} - t} \right)\\
 \Leftrightarrow {m_1}{c_1}\left( {60 - 20} \right) + mc\left( {60 - 20} \right) = 2mc\left( {100 - 60} \right)\\
 \Leftrightarrow 40{m_1}{c_1} + 40mc = 80mc\\
 \Leftrightarrow 40{m_1}{c_1} = 40mc\\
 \Leftrightarrow {m_1}{c_1} = mc\\
{Q_{thu}} = {Q_{toa}}\\
 \Leftrightarrow {m_1}{c_1}\left( {t' - {t_1}} \right) = 2mc\left( {{t_3} - t'} \right)\\
 \Leftrightarrow t' - 20 = 2\left( {100 - t'} \right)\\
 \Leftrightarrow t' - 20 = 200 - 2t'\\
 \Leftrightarrow 3t' = 220 \Rightarrow t' = 73,{33^o}C
\end{array}$

m1c1(t−t1)+mc(t−t1)=2mc(t3−t)⇔m1c1(60−20)+mc(60−20)=2mc(100−60)⇔40m1c1+40mc=80mc⇔40m1c1=40mc⇔m1c1=mcQthu=Qtoa⇔m1c1(t′−t1)=2mc(t3−t′)⇔t′−20=2(100−t′)⇔t′−20=200−2t′⇔3t′=220⇒t′=73,33oCQthu=Qtoa⇔m1c1(t−t1)+mc(t−t1)=2mc(t3−t)⇔m1c1(60−20)+mc(60−20)=2mc(100−60)⇔40m1c1+40mc=80mc⇔40m1c1=40mc⇔m1c1=mcQthu=Qtoa⇔m1c1(t′−t1)=2mc(t3−t′)⇔t′−20=2(100−t′)⇔t′−20=200−2t′⇔3t′=220⇒t′=73,33oC