Answer
a. $1.6\times10^{17}K$.
b. This is in the electroweak era.
Work Step by Step
a. Use the temperature–kinetic energy relationship $kT=KE$, from section 33–7. Find the temperature.
$$T=\frac{KE}{k}=\frac{(14\times10^{12}eV)(1.60\times10^{-19}J/eV)}{1.38\times10^{-23}J/K}=1.6\times10^{17}K$$
This is extraordinarily hot.
b. Looking at Figure 33–29, a temperature of $1.6\times10^{17}K$ is in the electroweak era.
