Keith Williams wrote:

Assume:
a 20' x 50' driveway = 1000sq.ft.
6" snow = ~1" water (frozen) = 170cu.ft. ice

.... 20x50x1/12 = 83 ft^3.

63 lbs/cu ft. = 1E4 lb

5250 lb (actually less, since ice is less dense than water.)

450 gm / lb. = 4.7E6 g
Latent heat of freezing 80cal/g = 3.7E8 cal
1 cal = .004 BTU = 1.5E6 BTU

.... 5250x144 = 756K Btu, eg 756K/(55-32) = 32869 lb (522 ft^3) of groundwater
cooling from 55 to 32 F.

Now where are you going to put 700ish gallons of water that's just
ready to freeze again? ;-)

Back in the 4' deep x 50' long x 522/50/4 = 2.6' wide stone-filled trench
with an EPDM liner on one side of the driveway, where it can warm up to
55 F and get pumped over the driveway to melt the next batch of snow :-)

Nick