Đáp án:

$$t=1944s=32,4phut$$

Giải thích các bước giải:

$$\begin{align}
  & {{m}_{1}}=0,5kg;{{t}_{0}}={{20}^{0}}C;t=3p=180s;{{t}_{1}}={{30}^{0}}C; \\ 
 & {{m}_{2}}=1kg;t=3p=180s;{{t}_{2}}={{28}^{0}}C; \\ 
 & {{m}_{3}}=1,5kg;{{t}_{3}}={{92}^{0}}C \\ 
\end{align}$$

Nhiệt lượng

$$\begin{align}
  & {{Q}_{1}}={{m}_{1}}.c.({{t}_{1}}-{{t}_{0}})+{{m}_{am}}.{{c}_{am}}.({{t}_{1}}-{{t}_{0}})=0,5.c.(30-20)+{{m}_{am}}.{{c}_{am}}.10=5c+10.{{m}_{am}}.{{c}_{am}}(J) \\ 
 & {{Q}_{2}}={{m}_{2}}.c.({{t}_{2}}-{{t}_{0}})+{{m}_{am}}.{{c}_{am}}.({{t}_{2}}-{{t}_{0}})=1.c.(28-20)+{{m}_{am}}.{{c}_{am}}.8=8c+8{{m}_{am}}.{{c}_{am}}(J) \\ 
 & {{Q}_{toa}}=P.t=180P(J) \\ 
\end{align}$$

khi cân bằng nhiệt

$$\begin{align}
  & 5c+10.{{m}_{am}}.{{c}_{am}}=180P \\ 
 & 8c+8{{m}_{am}}.{{c}_{am}}=180P \\ 
 & \Rightarrow c=9P;{{m}_{am}}.{{c}_{am}}=13,5P \\ 
\end{align}$$

khi khối lượng tăng lên 92 độ C

$$\begin{align}
  & t.P={{m}_{3}}.c.({{t}_{3}}-{{t}_{0}})+{{m}_{am}}.{{c}_{am}}.({{t}_{3}}-{{t}_{0}}) \\ 
 & \Leftrightarrow t.P=1,5.9P.(92-20)+13,5P.(92-20) \\ 
 & \Rightarrow t=1944s=32,4phut \\ 
\end{align}$$