Solution:

To have the same rms speed for \(\text{H}_2\) and \(\text{N}_2\), we use the formula for rms speed:

\[
v_{\text{rms}} = \sqrt{\frac{3k_B T}{m}}
\]

Since the rms speeds are equal, we can set up the equation:

\[
\sqrt{\frac{3k_B T_{\text{H}_2}}{m_{\text{H}_2}}} = \sqrt{\frac{3k_B T_{\text{N}_2}}{m_{\text{N}_2}}}
\]

Square both sides and simplify:

\[
\frac{T_{\text{H}_2}}{m_{\text{H}_2}} = \frac{T_{\text{N}_2}}{m_{\text{N}_2}}
\]

Since \(\text{N}_2\) is 14 times heavier than \(\text{H}_2\), we have \( m_{\text{N}_2} = 14 \, m_{\text{H}_2} \) and \( T_{\text{N}_2} = 27^\circ \text{C} = 300 \, \text{K} \).

Now solve for \( T_{\text{H}_2} \):

\[
T_{\text{H}_2} = \frac{T_{\text{N}_2}}{14} = \frac{300}{14} \approx 21.4 \, \text{K}
\]

So, the temperature at which \(\text{H}_2\) has the same rms speed as \(\text{N}_2\) at 27°C is approximately \( 21.4 \, \text{K} \).