The buoyant force in kg of a hot air balloon, at typical altitudes and
atmospheric pressure, is
V * (1013 / 2.87) * (1/tempAmbient - 1/tempEnvelope)
where V is in cubic meters and the temperatures are in degrees K (i.e., C + 273).
For example, when V = 1, tempAmbient = 20 deg C, tempEnvelope = 60 deg C,
then the buoyant force is 0.145 kg.
How much energy does it take to heat 1 cubic meter of air from 20 to 60 degrees?
Its specific heat is about 1.004 kJ/kg.K,
and 1 m^3 of air is 1.225 kg,
so the energy is 1.004 * 1.225 * (60-20) = 49.2 kJ.
A battery stores about 460 kJ/kg, so the amount of battery needed to
store 49.2 kJ is 49.2/460 = 0.11 kg.
That's just less than 0.145 kg, so if the rest of the balloon is under 35 grams
and if you're lucky, the balloon could just get airborne before the battery completely discharged.
Varying V doesn't change this analysis because both the buoyant force and the battery weight vary linearly with V. But let's change the difference in temperature from 60-20 to 25-20 degrees.
Buoyant force = 0.020 kg.
1.004 * 1.225 * (25-20) = 6.15 kJ energy, so
the battery weighs 6.15/460 = 0.013 kg.
It'll float if the rest of the balloon is under 7 grams. Not likely.
OK, let's go the other way, to 90-20. (Ouch, that's hot!)
Buoyant force = 0.232 kg.
1.004 * 1.225 * (90-20) = 86.1 kJ.
The battery weighs 86.1/460 = 0.187 kg.
It'll float if the rest of the balloon is under 45 grams.
So, the hotter the envelope, the more likely it'll float.
But even a "30C" rated battery takes 2 minutes to discharge safely.
That's slow enough for the envelope to lose heat to the outside.
I don't know how fast heat is lost across the envelope.
But even if you cheat by preheating before launch,
the battery might discharge too slowly to overcome that cooling.
(Could we exploit the battery's own "waste heat" as it approaches 60 deg C? Its specific heat is about 0.85 kJ/kg.K.
So for the 90 deg C scenario,
this extra thermal energy is 0.85 * 0.187 * (60 - 20) = 6.4 kJ. Just 7% more than the original 86.1 kJ, not enough to overcome the weight increase due to the extra fan, fins, etc.)
Achieving liftoff for a few seconds would be barely possible. So you might as well leave the battery on the ground, tethered to the balloon.