Đáp án:

$\begin{align}
  & a){{c}_{1}}=880J/kg.K \\ 
 & b){{t}_{cb}}=77,{{85}^{0}}C \\ 
\end{align}$

Giải thích các bước giải:

$M=0,4kg;{{t}_{1}}={{25}^{0}}C;m=0,1kg;{{t}_{2}}={{100}^{0}}C;{{t}_{3}}=65,{{8}^{0}}C$

a) khi có sự cân bằng nhiêt

$\begin{align}
  & {{Q}_{1}}={{Q}_{2}} \\ 
 & M.{{c}_{1}}.({{t}_{3}}-{{t}_{1}})=m.{{c}_{2}}.({{t}_{2}}-{{t}_{3}}) \\ 
 & \Rightarrow {{c}_{1}}=\dfrac{m.{{c}_{2}}.({{t}_{2}}-{{t}_{3}})}{M.({{t}_{3}}-{{t}_{1}})} \\ 
 & =\dfrac{0,1.4200.(100-65,8)}{0,4.(65,8-25)} \\ 
 & =880J/kg.K \\ 
\end{align}$

d) m'=0,1kg

$\begin{align}
  & {{Q}_{thu}}={{Q}_{toa}} \\ 
 & (M.{{c}_{1}}+m.{{c}_{2}}).({{t}_{cb}}-{{t}_{3}})=m'.{{c}_{2}}.({{t}_{2}}-{{t}_{cb}}) \\ 
 & \Leftrightarrow (0,4.880+0,1.4200).({{t}_{cb}}-65,8)=0,1.4200.(100-{{t}_{cb}}) \\ 
 & \Rightarrow {{t}_{cb}}=77,{{85}^{0}}C \\ 
\end{align}$