Đáp án:
a) 4
b) \({\dfrac{{90}}{7}^o}C\)
Giải thích các bước giải:
a)
* Phương trình cân bằng nhiệt khi rót lần 1 là:
\(\begin{array}{l}
{m_0}{c_0}\Delta {t_1} = mc\left( {t - {t_1}} \right)\\
\Rightarrow {m_0}{c_0}\Delta {t_1} = mc\left( {t - {t_0} - \Delta {t_1}} \right)\\
\Rightarrow \Delta {t_1}\left( {{m_0}{c_0} + mc} \right) = mc\left( {t - {t_0}} \right)\\
\Rightarrow \Delta {t_1} = \dfrac{{mc}}{{{m_0}{c_0} + mc}}\left( {t - {t_0}} \right)
\end{array}\)
* Phương trình cân bằng nhiệt khi rót lần 2 là:
\(\begin{array}{l}
{m_0}{c_0}\Delta {t_2} = 2mc\left( {t - {t_2}} \right)\\
\Rightarrow {m_0}{c_0}\Delta {t_2} = 2mc\left( {t - {t_0} - \Delta {t_2}} \right)\\
\Rightarrow \Delta {t_2}\left( {{m_0}{c_0} + 2mc} \right) = 2mc\left( {t - {t_0}} \right)\\
\Rightarrow \Delta {t_2} = \dfrac{{2mc}}{{{m_0}{c_0} + 2mc}}\left( {t - {t_0}} \right)
\end{array}\)
Suy ra:
\(\begin{array}{l}
\dfrac{{\Delta {t_1}}}{{\Delta {t_2}}} = \dfrac{{\dfrac{{mc}}{{{m_0}{c_0} + mc}}\left( {t - {t_0}} \right)}}{{\dfrac{{2mc}}{{{m_0}{c_0} + 2mc}}\left( {t - {t_0}} \right)}} = \dfrac{1}{2}.\dfrac{{{m_0}{c_0} + 2mc}}{{{m_0}{c_0} + mc}}\\
\Rightarrow \dfrac{6}{{6 + 4}} = \dfrac{1}{2}.\dfrac{{{m_0}{c_0} + 2mc}}{{{m_0}{c_0} + mc}}\\
\Rightarrow \dfrac{6}{5} = \dfrac{{{m_0}{c_0} + 2mc}}{{{m_0}{c_0} + mc}}\\
\Rightarrow 6\left( {{m_0}{c_0} + mc} \right) = 5\left( {{m_0}{c_0} + 2mc} \right)\\
\Rightarrow {m_0}{c_0} = 4mc
\end{array}\)
b) * Khi rót lần 3, ta có:
\(\begin{array}{l}
\Delta {t_3} = \dfrac{{3mc}}{{{m_0}{c_0} + 3mc}}\left( {t - {t_0}} \right)\\
\Rightarrow \dfrac{{\Delta {t_1}}}{{\Delta {t_3}}} = \dfrac{{\dfrac{{mc}}{{{m_0}{c_0} + mc}}\left( {t - {t_0}} \right)}}{{\dfrac{{3mc}}{{{m_0}{c_0} + 3mc}}\left( {t - {t_0}} \right)}}\\
\Rightarrow \dfrac{{\Delta {t_1}}}{{\Delta {t_3}}} = \dfrac{1}{3}.\dfrac{{{m_0}{c_0} + 3mc}}{{{m_0}{c_0} + mc}} = \dfrac{1}{3}.\dfrac{{4mc + 3mc}}{{4mc + mc}}\\
\Rightarrow \dfrac{6}{{\Delta {t_3}}} = \dfrac{1}{3}.\dfrac{7}{5}\\
\Rightarrow \Delta {t_3} = {\dfrac{{90}}{7}^o}C
\end{array}\)
