@criver you are making the mistake of air temp = steel temp. Air is a very good insulator so it heats up more when electricity passes through it.
The tanks they are using are Type-90s with composite armor 320mm of thickness so lets say a third or 110mm or 11cm is steel and .5m or 50cm wide cheek and tall(the flat not moving bit). For a cross section area 2500cm2 or 0.25m2 with a volume of 27500 cm3 or 0.0275m3 (this is melting through the armor)
Let us assume this is MIL-A-46100 steel which is used stop hyper-velocity projectiles like a tank round. This is Iron/Chromium/Molybdenum alloy and can say it has a density of 7.78g/cm3
Iron/Chromium/Molybdenum alloy on my chart says it has a resistivity of 49.9µΩcm or 4.99*10^-7Ωm at 20°C
Iron/Chromium/Molybdenum does not have a melting point listed(I wonder why
) so lets go with Iron/Molybdenum temp of 1452°C and round it to 1450°C.
Specific heat is not listed(I wonder why
) but iron based steel tend to be about 500J/°K/kg or 0.5j/°C/g/
Let us do some math
There will be (27500cm3)(7.78g/cm3) = 213950g or 213.95kg of steel that needs to be melted.
The steel's resistance is resistivity * distance / cross-sectional area
(14.99*10^-7 Ωm)(.11 m)/(.25 m2) = 6.5956×10^-7 Ω or 0.65956 µΩ
I will be nice and say the lightning will get is peak current of 400kA from positive lightning which is about 5% of lightning instead of the more common 30kA average which can peak up to 120kA of the other 95%. We will also say the main stroke aka the part that is most powerful takes 50ms as no one agrees but 50 is in the middle. The follow up strokes require a lot less power as plasma is far more conductive than air.
Positive
(400 kA)²(0.65956 µΩ) = 0.10553 MW
(0.10553 MW)(1 ms) = 0.10553 kJ
Negative peak
(120 kA)²(0.65956 µΩ) = 0.00949766 MW
(0.00949766 MW)(1 ms) = 0.00949766 kJ
Negative Average
(30kA)²(0.65956 µΩ) = 0.000593604 MW
(0.000593604 MW)(1 ms) = 0.000593604 kJ
so to find out how long it will take to melt it we need to do some algebra making (power/(Specific heat * mass)=temperature rose)
Positive
(0.10553 kJ)/((0.5j/°C/g)(213950g))= 9.86×10^-4 °C (Celsius degrees difference)
(1450°C)/(9.86×10^-4 °C) = 1471 seconds or 24.5 minutes
Negative peak
(0.00949766 kJ)/((0.5j/°C/g)(213950g))= 9.86×10^-4 °C (Celsius degrees difference)
(1450°C)/(8.88×10^-5) = 16329 seconds or 4 hours and 32 minutes rounded
Negative Average
(0.000593604 kJ)/((0.5j/°C/g)(213950g))= 5.55×10^-6 °C (Celsius degrees difference)
(1450°C)/5.55×10^-6 °C = 261261 seconds or 3 days 34 minutes rounded
Calculating amount of lightning strikes is pointless just multiply the seconds by 20
@GambierBay said 12 hours because that is the time it takes a industrial electric furnace to melt tonnes(1000 Kg) of steel (in the tonnes range) and you want to do it with a lightning bolt.