Think you're somewhat off. Quaise says they intend to allow their tech to be "plugged into existing oil rigs," and most boreholes are 5-20 in wide.
20 in down 7.6 mi is 0.2 m^2 down 12321 m. This gives 2500 m^3 or 2.5e-6 km^3 of rock.
Google says "rocks are generally between 1600 kg/m^3 (sediments) and 3500 kg/m^3 (gabbro)" so going with 2500 km/m^3 that would be 6240000 kg, 6.24 Gg, or 6.24 thousand metric tons.
Heat of vaporization of water is roughly 2250 kJ/kg, or 537 kcal/kg, so if the mass was water it would need 14e9 kJ, 14 TJ, or 3.35e9 kcal to vaporize.
For a 5 in borehole divide all the above by 16. This ignores the heat required to actually change the temperature of the water, but I'm betting you can get a lot of that back via condensation, and for this reason I'd also assume that much of the rock will not end up in atmosphere. These assumptions also go entirely out the window of there's an inflow of material, but unfortunately I'm betting that's exactly what will happen.
Edit: Fixed post after I confused radius and diameter. Someone should probably check my math as well.