Answer
408.94 K = 135.96 Degree Celsius
Work Step by Step
Let $T_{1} = 15^{o}C + 273 K = 288 K$, $V_{1} = 2.5 L$, $V_{2} = 3550 mL = 3.55 L$, $T_{2} = ?$
Using Charles's Law, we have:
$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$
$\frac{2.5 L}{288 K} = \frac{3.55 L}{T_{2}}$
$T_{2} = \frac{3.55 L}{2.5 L}$ x 288K
= 408.96 K