Answer
$120K$
Work Step by Step
We know that
$\Delta T=\frac{Q}{mc}$
$\implies \Delta T=\frac{\frac{1}{2}K.E}{mc}$
$\implies \Delta T=\frac{\frac{1}{2}(\frac{1}{2}mv^2)}{mc}$
This simplifies to:
$\Delta T=\frac{v^2}{4c}$
We plug in the known values to oobtain:
$\Delta T=\frac{(250)^2}{4(128)}=120K$
