## College Physics (4th Edition)

$T = 2220~K$
We can find the temperature: $\frac{3}{2}~k~T = \overline{KE}$ $T = \frac{2~\overline{KE}}{3~k}$ $T = \frac{(2)(4.60\times 10^{-20}~J)}{(3)(1.38\times 10^{-23}~J/K)}$ $T = 2220~K$