Đáp án:
5 lần
Giải thích các bước giải:
lần 1:
\(\begin{array}{*{35}{l}}
\text{ }\!\!~\!\!\text{ } & {{m}_{tr}}.c.({{t}_{1}}-{{t}_{0}})={{m}_{s}}.c.(60-{{t}_{1}}) \\
\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ } & \Leftrightarrow 0,2.({{t}_{1}}-10)=0,05.(60-{{t}_{1}}) \\
\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ } & \Rightarrow {{t}_{1}}={{20}^{0}}C \\
\text{ }\!\!~\!\!\text{ } & {} \\
\end{array}\)
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rót vào lần 2:
\(\begin{array}{*{35}{l}}
\text{ }\!\!~\!\!\text{ } & 0,2.c.({{t}_{2}}-{{t}_{1}})=0,05.c.(60-{{t}_{2}}) \\
\text{ }\!\!~\!\!\text{ } & \Leftrightarrow 0,2.({{t}_{2}}-20)=0,05.(60-{{t}_{2}}) \\
{} & \Rightarrow {{t}_{2}}={{28}^{0}}C \\
\text{ }\!\!~\!\!\text{ } & {} \\
\end{array}\)
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=> rót thêm lần 3:
\(\begin{array}{*{35}{l}}
\text{ }\!\!~\!\!\text{ } & 0,2.c.({{t}_{3}}-{{t}_{2}})=0,05.c.(60-{{t}_{3}}) \\
\text{ }\!\!~\!\!\text{ } & \Leftrightarrow 0,2.({{t}_{3}}-28)=0,05.(60-{{t}_{3}}) \\
\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ } & \Rightarrow {{t}_{3}}=34,{{4}^{0}}C \\
\text{ }\!\!~\!\!\text{ } & {} \\
\end{array}\)
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=> rót thêm lần 4:
\(\begin{array}{*{35}{l}}
\text{ }\!\!~\!\!\text{ } & 0,2.c.({{t}_{4}}-{{t}_{3}})=0,05.c.(60-{{t}_{4}}) \\
\text{ }\!\!~\!\!\text{ } & \Leftrightarrow 0,2.({{t}_{4}}-34,4)=0,05.(60-{{t}_{4}}) \\
\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ } & \Rightarrow {{t}_{4}}=39,{{52}^{0}}C \\
\text{ }\!\!~\!\!\text{ } & {} \\
\end{array}\)
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=> rót thêm lần 5:
\(\begin{array}{*{35}{l}}
\text{ }\!\!~\!\!\text{ } & 0,2.c.({{t}_{5}}-{{t}_{4}})=0,05.c.(60-{{t}_{5}}) \\
\text{ }\!\!~\!\!\text{ } & \Leftrightarrow 0,2.({{t}_{5}}-39,52)=0,05.(60-{{t}_{5}}) \\
\text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ } & \Rightarrow {{t}_{5}}=43,{{616}^{0}}C \\
\text{ }\!\!~\!\!\text{ } & {} \\
\end{array}\)
