Đáp án:

 $\begin{align}
  & 1){{c}_{KL}}=131,25J/kg.K \\ 
 & 2){{t}_{cb}}={{18}^{0}}C \\ 
 & 3){{m}_{dong}}=1,58kg \\ 
 & 4){{t}_{cb}}=28,{{7}^{0}}C \\ 
\end{align}$

Giải thích các bước giải:

 Bài 1:

${{m}_{KL}}=0,6kg;{{t}_{KL}}={{100}^{0}}C;{{m}_{nc}}=0,5kg;{{t}_{nc}}=58,{{5}^{0}}C;{{t}_{cb}}={{80}^{0}}C$

khi cân bằng nhiêt xảy ra:

$\begin{align}
  & {{Q}_{toa}}={{Q}_{thu}} \\ 
 & {{m}_{KL}}.{{c}_{KL}}.({{t}_{KL}}-{{t}_{cb}})={{m}_{nc}}.{{c}_{nc}}.({{t}_{cb}}-{{t}_{nc}}) \\ 
 & \Leftrightarrow 0,6.{{c}_{KL}}.(100-60)=0,5.4200.(60-58,5) \\ 
 & \Rightarrow {{c}_{KL}}=131,25J/kg.K \\ 
\end{align}$

Bài 2:

${{m}_{nc}}=1,2lit=1,2kg;{{t}_{nc}}={{15}^{0}}C;{{m}_{dong}}=0,5kg;{{t}_{dong}}={{100}^{0}}C;$

khi cân bằng nhiệt xảy ra:

$\begin{align}
  & {{Q}_{toa}}={{Q}_{thu}} \\ 
 & \Leftrightarrow {{m}_{dong}}.{{c}_{dong}}.({{t}_{dong}}-{{t}_{cb}})={{m}_{nc}}.{{c}_{nc}}.({{t}_{cb}}-{{t}_{nc}}) \\ 
 & \Leftrightarrow 0,5.380.(100-{{t}_{cb}})=1,2.4200.({{t}_{cb}}-15) \\ 
 & \Rightarrow {{t}_{cb}}={{18}^{0}}C \\ 
\end{align}$

Bài 3:

${{m}_{nc}}=5lit=5kg;{{t}_{0}}={{100}^{0}}C;{{t}_{nguoi}}={{25}^{0}};{{Q}_{toa}}=1629kJ$

nhiệt lượng tỏa ra:

$\begin{align}
  & {{Q}_{toa}}={{Q}_{dong}}+{{Q}_{nc}} \\ 
 & \Leftrightarrow {{1620.10}^{3}}=({{m}_{d}}.{{c}_{d}}+{{m}_{nc}}.{{c}_{nc}}).({{t}_{0}}-{{t}_{nguoi}}) \\ 
 & \Leftrightarrow {{1620.10}^{3}}=({{m}_{d}}.380+5.4200).(100-25) \\ 
 & \Rightarrow {{m}_{d}}=1,58kg \\ 
\end{align}$

Bài 4:

${{m}_{dong}}=0,6kg;{{t}_{dong}}={{85}^{0}}C;{{m}_{nc}}=0,35kg;{{t}_{nc}}={{20}^{0}}C$

Khi có sự cân bằng nhiệt:

$\begin{align}
  & {{Q}_{toa}}={{Q}_{thu}} \\ 
 & \Leftrightarrow {{m}_{d}}.{{c}_{d}}.({{t}_{d}}-{{t}_{cb}})={{m}_{nc}}.{{c}_{nc}}.({{t}_{cb}}-{{t}_{nc}}) \\ 
 & \Leftrightarrow 0,6.380.(85-{{t}_{cb}})=0.35.4200.({{t}_{cb}}-20) \\ 
 & \Rightarrow {{t}_{cb}}=28,{{7}^{0}}C \\ 
\end{align}$