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Answer 1 · 2017-06-01T14:15:23

$37.5^(@)$ $"C"$

Let's use Charles' law $frac(V_(1))(T_(1)) = frac(V_(2))(T_(2))$ :

$Rightarrow frac(100.0 " ml")(15.0^(@) " C") = frac(250.0 " ml")(T_(2))$

Dividing both sides of the equation by $100.0$ $"ml"$ :

$Rightarrow frac(1)(15.0^(@) " C") = frac(2.5)(T_(2))$

$Rightarrow T_(2) = 2.5 times 15.0^(@)$ $"C"$

$therefore T_(2) = 37.5^(@)$ $"C"$

Therefore, the final temperature will be $37.5^(@)$ $"C"$ .