I'll do my best to formulate this as a purely electrical problem so that it isn't claimed to be belonging to Vehicle Maintenance SE.

We have a lead-acid starter battery (not a deep cycle fashion) that has been sitting in low temperature conditions - something like -20 degrees Celsius (-4 Fahrenheit) and so it's capacity has seriously reduced compared to warmer weather. We have an engine starter that drains about 200 amperes from the battery for several seconds.

Now there's a claim that simply trying the start the engine is not the best practice because the very cold battery will not give much power and the engine might not start. The "better practice" is claimed to be to turn the headlights on for about half a minute so that "the juice starts flowing" and the battery warms up and outputs more current.

So the myth goes. Now a typical headlight bulb power is 55 watts, so for both headlights we get 110 watts which means like 9 amperes current through the battery.

A typical car battery weighs about 17 kilograms (about 40 pounds). So the claim basically means that 9 amperes for half a minute is enough to warm a cooled to -20 Celsius lead acid battery and notably improve its characteristic.

Then shouldn't draining about 200 amperes for several seconds make the battery heat further and perhaps boil and explode?

From the point of how a typical lead acid starter battery works at about -20 Celsius is it possible that running about 9 amperes load for about half a minute would help it warm and output more power and get the engine starter run better?

  • \$\begingroup\$ There's an assumption here that any mechanism behind a possible beneficial effect would be thermal; I wonder if that is the only possibility? \$\endgroup\$ Commented Jan 30, 2014 at 6:10
  • \$\begingroup\$ @ChrisStratton: Interesting thought. Indeed it could be possible that some kind of "juice flowing" effect exists for batteries. \$\endgroup\$
    – sharptooth
    Commented Jan 30, 2014 at 6:13

1 Answer 1


Battery heating can be roughly defined using its internal resistance. We can get a rough estimate of a battery's internal resistance using it's cold cranking amp rating. This is rated at -18C, which is close to -20C.

For a nominal 12V battery: $$ R = \frac{12V - 7.2V}{CCA} $$

Let's assume this is constant vs. current draw.

Let's assume at time t=0, the battery is completely insulated from the world and has a heat capacity similar to pure water.

The amount of energy required to heat 17 kg of water from say even -20C to -18C is: $$ E = 4.2 \frac{J}{g K} \cdot 17 kg \cdot (20^oC - 18^oC) = 142.8 kJ $$

For a 9A draw, this would take:

$$ t = \frac{E}{(9A)^2 \cdot R} $$

Suppose you have a 200 CCA battery. Then we can find t = 73 seconds.

Problems with this analysis: Obviously the full weight of the battery is not just the liquid inside. Thus the time required to heat the battery 2 degrees C would be less. However, a larger problem is that you probably don't have a perfectly insulated battery. The 9A draw only produces ~2W, which could easily be lost from the battery (the cables are extremely good conductors away from the battery).

At temperatures much colder you're going to have to start worrying about fluids freezing, thus many things could just stop working. You're much better off following proper car winterization guide.

  • \$\begingroup\$ I like the approach, but where did you get the 7.2V term? \$\endgroup\$
    – Phil Frost
    Commented Jan 29, 2014 at 13:42
  • 1
    \$\begingroup\$ It seems to be from the definition of "cold cranking amps", specifically maintaining 1.2v per (nominally 2v) cell. \$\endgroup\$ Commented Jan 29, 2014 at 16:31
  • 4
    \$\begingroup\$ Does this get any better if you remove the headlight bulbs and hold them against the battery to warm it up directly? ;) \$\endgroup\$
    – John U
    Commented Jan 29, 2014 at 17:36
  • \$\begingroup\$ Most of the weight of the battery is lead, but the goal isn't necessarily to heat the bulk of the lead, but rather to heat the surface layer. I would expect that layer would with usage get warmer than the rest of the cell, but I don't know how quickly it would lose that heat back to the bulk of the lead. \$\endgroup\$
    – supercat
    Commented Aug 6, 2014 at 23:32

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