Why is 1,3-pentadiene favored in the reaction of pent-3-en-2-ol with sulfuric acid?
Hydrogenation of 1,trans-3-pentadiene (your top molecule) and the allene, penta-2,3-diene (your bottom molecule), both produce pentane plus heat.  Heat is given off because both of the dienes plus hydrogen is less stable than pentane.  With 1,trans-3-pentadiene, 53.1 kcal/m of heat is given off; with penta-2,3-diene 66.8 kcal/m of heat is given off (data from Frances Maron Fraser and Edward J. Prosen, Journal of Research of the National Bureau of Standards Vol. 54, No. 3, March 1955 Research Paper 2575, p. 143-148). Clearly the allene is of higher energy than the 1,3-diene by about 15.7 kcal/m. 

The reaction conditions (strong acid and heat) suggest that an equilibrium mixture of possible products will result.  Around room temperature, each 1.4 kcal/m difference in energy between two isomers will shift the equilibrium between them by a power of 10 ($\ce{\Delta G=-RTlnK}$).  A 15 kcal/m energy difference between your two dienes suggests that the equilibrium will be ~ $\mathrm{10^{10}:1}$ favoring the 1,3-pentadiene. If any of the allene forms, it will rapidly convert to the 1,3-pentadiene.

Side Note: Why are cumulenes generally less stable than 1,3-dienes?


Conjugation which results in resonance stabilization is present in the 1,3-diene but not in the 2,3-diene.
Allenes (in fact, all cumulenes) have considerable strain built into them. If you view a double bond as a two-membered ring, it is easy to see this.  The bond angle in the two-membered ring is severely pinched down.  In turn, the bond angle external to the double bond tries to open up and relieve strain.  This strain relief, can't happen with the central carbon in an allene because it has a double bond on both sides.