The mechanism is an $\bf E_1$ reaction in which the substrate is a protonated alcohol.
In this step, an acid–base reaction, a proton is rapidly transferred from the acid to one of the unshared electron pairs of the alcohol. In dilute sulfuric acid the acid is a hydronium ion; in concentrated sulfuric acid the initial proton donor is sulfuric acid itself. The leaving group is a molecule of water.
The carbon–oxygen bond breaks heterolytically. The bonding electrons depart with the
water molecule and leave behind a carbocation. The carbocation is, of course, highly reactive because the central carbon atom has only six electrons in its valence level, not eight.
Finally, in step 3, a water molecule removes a proton from the $\beta$ carbon of the carbocation. The result is the formation of a hydronium ion and an
In step 3, also an acid–base reaction, any one of the nine protons available at the three
methyl groups can be transferred to a molecule of water. The electron pair left behind when
a proton is removed becomes the second bond of the double bond of the alkene. Notice
that this step restores an octet of electrons to the central carbon atom.
Though water gives proton in step one and other molecule of water takes proton in last step, it is usually written as a produce of alkene and water until a mechanism is specifically mentioned.This is also reminding of the catalytic nature of water in this reaction.