# Question #7be8f

Jun 1, 2017

${37.5}^{\circ}$ $\text{C}$

#### Explanation:

Let's use Charles' law $\frac{{V}_{1}}{{T}_{1}} = \frac{{V}_{2}}{{T}_{2}}$:

$R i g h t a r r o w \frac{100.0 \text{ ml")(15.0^(@) " C") = frac(250.0 " ml}}{{T}_{2}}$

Dividing both sides of the equation by $100.0$ $\text{ml}$:

$R i g h t a r r o w \frac{1}{{15.0}^{\circ} \text{ C}} = \frac{2.5}{{T}_{2}}$

$R i g h t a r r o w {T}_{2} = 2.5 \times {15.0}^{\circ}$ $\text{C}$

$\therefore {T}_{2} = {37.5}^{\circ}$ $\text{C}$

Therefore, the final temperature will be ${37.5}^{\circ}$ $\text{C}$.