Gấp...................

Bài 4: 

Ta có: 

\({m_1} + {m_2} = m = 50g = 0,05kg\)

Phương trình cân bằng nhiệt:

\(\begin{array}{l}
\left( {{m_1}{c_1} + {m_2}{c_2}} \right)\left( {{t_1} - t} \right) = \left( {c + m'{c_0}} \right)\left( {t - {t_2}} \right)\\
 \Rightarrow \left( {130{m_1} + 380{m_2}} \right)\left( {136 - 18} \right) = \left( {30 + 0,1.4200} \right)\left( {18 - 14} \right)\\
 \Rightarrow 130{m_1} + 380{m_2} = \dfrac{{900}}{{59}}\\
 \Rightarrow \left\{ \begin{array}{l}
{m_1} = 0,015kg\\
{m_2} = 0,035kg
\end{array} \right.
\end{array}\)

Bài 5: 

Ta có: 

\(\begin{array}{l}
{t_1} - t = \dfrac{1}{2}\left( {{t_1} - {t_2}} \right)\\
 \Rightarrow t = {t_1} - \dfrac{1}{2}\left( {{t_1} - {t_2}} \right) = \dfrac{1}{2}{t_1} + {t_2}
\end{array}\)

Phương trình cân bằng nhiệt:

\(\begin{array}{l}
{m_1}{c_1}\left( {{t_1} - t} \right) = {m_2}{c_2}\left( {t - {t_2}} \right)\\
 \Rightarrow {m_1}{c_1}.\left( {{t_1} - \left( {\dfrac{1}{2}{t_1} + {t_2}} \right)} \right) = {m_2}{c_2}\left( {\dfrac{1}{2}{t_1} + {t_2} - {t_2}} \right)\\
 \Rightarrow {m_1}{c_1}.\left( {\dfrac{1}{2}{t_1} - {t_2}} \right) = {m_2}{c_2}.\dfrac{1}{2}{t_1}\\
 \Rightarrow \dfrac{{{m_1}}}{{{m_2}}} = \dfrac{{\dfrac{{{c_2}{t_1}}}{2}}}{{{c_1}.\dfrac{{{t_1} - 2{t_2}}}{2}}} = \dfrac{{{c_2}{t_1}}}{{{c_1}\left( {{t_1} - 2{t_2}} \right)}}
\end{array}\)

Bài 6: 

Ta có: 

\({m_1} + {m_2} = m = 50g = 0,05kg\)

Phương trình cân bằng nhiệt:

\(\begin{array}{l}
\left( {{m_1}{c_1} + {m_2}{c_2}} \right)\left( {{t_1} - t} \right) = \left( {q + m'{c_0}} \right)\left( {t - {t_2}} \right)\\
 \Rightarrow \left( {377{m_1} + 126{m_2}} \right)\left( {136 - 18} \right) = \left( {50 + 0,1.4200} \right)\left( {18 - 14} \right)\\
 \Rightarrow 377{m_1} + 126{m_2} = \dfrac{{940}}{{59}}\\
 \Rightarrow \left\{ \begin{array}{l}
{m_1} = 0,038kg\\
{m_2} = 0,012kg
\end{array} \right.
\end{array}\)