Answer
(a)
$$T=6 \times 10^{12} \mathrm{K}$$
From Fig. $33-29,$ this corresponds to a time of $\left[\approx 10^{-5} \mathrm{s}\right]$
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(b) $$T=1 \times 10^{14} \mathrm{K}$$
From Fig. $33-29,$ this corresponds to a time of $\left[\approx 10^{-7} \mathrm{s}\right]$
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(c)
$$T=1 \times 10^{12} \mathrm{K}$$
From Fig. $33-29,$ this corresponds to a time of $\left[\approx 10^{-4} \mathrm{s}\right]$
please note that may be exist some variation in the answers due to differences in reading the figure.
Work Step by Step
We approximate the temperature-energy relationship by $k T=E=m c^{2}$
$$
k T=m c^{2} \rightarrow T=\frac{m c^{2}}{k}
$$
(a)
$$T=\frac{m c^{2}}{k}=\frac{\left(500 \mathrm{MeV} / c^{2}\right) c^{2}\left(1.60 \times 10^{-13} \mathrm{J} / \mathrm{MeV}\right)}{1.38 \times 10^{-23} \mathrm{J} / \mathrm{K}}=6 \times 10^{12} \mathrm{K}$$
From Fig. $33-29,$ this corresponds to a time of $\left[\approx 10^{-5} \mathrm{s}\right]$
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(b) $$T=\frac{m c^{2}}{k}=\frac{\left(9500 \mathrm{MeV} / c^{2}\right) c^{2}\left(1.60 \times 10^{-13} \mathrm{J} / \mathrm{MeV}\right)}{1.38 \times 10^{-23} \mathrm{J} / \mathrm{K}}=1 \times 10^{14} \mathrm{K}$$
From Fig. $33-29,$ this corresponds to a time of $\left[\approx 10^{-7} \mathrm{s}\right]$
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(c)
$$T=\frac{m c^{2}}{k}=\frac{\left(100 \mathrm{MeV} / c^{2}\right) c^{2}\left(1.60 \times 10^{-13} \mathrm{J} / \mathrm{MeV}\right)}{1.38 \times 10^{-23} \mathrm{J} / \mathrm{K}}=1 \times 10^{12} \mathrm{K}$$
From Fig. $33-29,$ this corresponds to a time of $\left[\approx 10^{-4} \mathrm{s}\right]$
$\text{please note that may be exist some variation in the answers due to differences}$ $\text{in reading the figure.}$
