Đáp án:

a) \( - {40^o}C\)

b)  \({58^o}C\)

Giải thích các bước giải:

a) * Giai đoạn 1:

Ta có: 

\(\begin{array}{l}
{q_1}\left( {{t_2} - {t_1}} \right) = P{t_1}\\
 \Rightarrow {q_1}\left( {80 - 40} \right) = P.2.60\\
 \Rightarrow {q_1} = 3P
\end{array}\)

* Giai đoạn 2:

Ta có:

\(\begin{array}{l}
{q_1}\left( {{t_2} - {t_1}} \right) = {q_2}\left( {{t_1} - {t_0}} \right)\\
 \Rightarrow 3P\left( {80 - 40} \right) = {q_2}\left( {40 - {t_0}} \right)\\
 \Rightarrow 120P = {q_2}\left( {40 - {t_0}} \right)
\end{array}\)

* Giai đoạn 3:

Ta có:

\(\begin{array}{l}
\left( {{q_1} + {q_2}} \right)\left( {{t_2} - {t_1}} \right) = P{t_2}\\
 \Rightarrow \left( {3P + \dfrac{{120P}}{{40 - {t_0}}}} \right).\left( {80 - 40} \right) = P.3.60\\
 \Rightarrow \left( {3 + \dfrac{{120}}{{40 - {t_0}}}} \right).40 = 180\\
 \Rightarrow {t_0} =  - {40^o}C
\end{array}\)

b) * Giai đoạn 4:

Ta có:

\(\begin{array}{l}
{q_2} = \dfrac{{120P}}{{40 - {t_0}}} = \dfrac{{120P}}{{40 - \left( { - 40} \right)}} = 1,5P\\
\left( {{q_1} + {q_2}} \right)\left( {100 - {t_1}} \right) = {q_3}\left( {{t_1} - {t_0}} \right)\\
 \Rightarrow \left( {3P + 1,5P} \right).\left( {100 - {t_1}} \right) = {q_3}\left( {{t_1} + 40} \right)\\
 \Rightarrow {q_3} = \dfrac{{4,5P\left( {100 - {t_1}} \right)}}{{{t_1} + 40}}
\end{array}\)

* Giai đoạn 5: 

Ta có:

\(\begin{array}{l}
\left( {{q_1} + {q_2} + {q_3}} \right)\left( {100 - {t_1}} \right) = P{t_3}\\
 \Rightarrow \left( {3P + 1,5P + \dfrac{{4,5P\left( {100 - {t_1}} \right)}}{{{t_1} + 40}}} \right)\left( {100 - {t_1}} \right) = P.\left( {11 - 6,5} \right).60\\
 \Rightarrow \left( {4,5 + \dfrac{{4,5\left( {100 - {t_1}} \right)}}{{{t_1} + 40}}} \right)\left( {100 - {t_1}} \right) = 270\\
 \Rightarrow {t_1} = {58^o}C
\end{array}\)