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Answer 1 · 2017-06-01T14:15:23

${37.5}^{\circ}$ $37.5^(@)$ $C$ $"C"$

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

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

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

$\Rightarrow \frac{1}{{15.0}^{\circ} C} = \frac{2.5}{T_{2}}$ $Rightarrow frac(1)(15.0^(@) " C") = frac(2.5)(T_(2))$

$\Rightarrow T_{2} = 2.5 \times {15.0}^{\circ}$ $Rightarrow T_(2) = 2.5 times 15.0^(@)$ $C$ $"C"$

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

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