Solution:

The change in length \( \Delta L \) due to temperature change is given by:

\[
\Delta L = L_0 \alpha \Delta T
\]

where:
- \( L_0 = 10 \, \text{cm} \)
- \( \alpha = 11 \times 10^{-6} / ^\circ \text{C} \)
- \( \Delta T = 20^\circ \text{C} - 19^\circ \text{C} = 1^\circ \text{C} \)

Substitute the values:

\[
\Delta L = 10 \times 11 \times 10^{-6} \times 1 = 11 \times 10^{-5} \, \text{cm}
\]

Thus, the bar will be \( 11 \times 10^{-5} \, \text{cm} \) shorter at 19ºC.