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Answer:
[tex]114.4^{\circ}\text{C}[/tex]
Explanation:
[tex]T_1[/tex] = Initial temperature = [tex]25^{\circ}\text{C}=25+273\ \text{K}[/tex]
[tex]V_1[/tex] = Initial volume = 50 L
[tex]T_2[/tex] = Final temperature
[tex]V_2[/tex] = Final volume = 65 L
We have the relation
[tex]\dfrac{V_2}{V_1}=\dfrac{T_2}{T_1}\\\Rightarrow T_2=\dfrac{V_2}{V_1}\times T_1\\\Rightarrow T_2=\dfrac{65}{50}\times (25+273)\\\Rightarrow T_2=387.595\ \text{K}=387.4-273\\\Rightarrow T_2=114.4^{\circ}\text{C}[/tex]
The final temperature of the system would be [tex]114.4^{\circ}\text{C}[/tex].