How long does it take to recharge a car battery after starting?

No, it isn't Joules in = Joules out. To a first approximation, it's Coulombs in = Coulombs out. It's the electrons flowing through the circuit that participate in the chemical reaction inside the battery (but not at 100% efficiency).

Forget about the energy/power/voltage calculation and just make the ampere-seconds for charging equal to the ampere-seconds for discharging, and then multiply by a fudge factor to account for the inefficiencies.

500A × 3s =1500 A-s = 2A × 750s = 10A × 150s

750s = 12.5 minutes

Figure about 90% efficiency, so the 12.5 minutes / 0.90 = about 14 minutes.

For "Close enough", you can use

T_charge = T_discharge * (i_discharge / i_charge) * k

k is a unitless current efficiency factor and varies with battery chemistry, charge and discharge rates, battery state of charge and phase of the moon (and sometimes whether today is a bank holiday), but for a

• lead acid battery: about 1.1 to 1.2
• lithium ion battery: about 1.01
• nickel-metal hydride (NiMH): about 1.15 to 1.2

This just says that charge and discharge times are inversely proportional to current drain multiplied by a variable constant.

The "constant" varies because of many factors. Lithium chemistries have no secondary reactions that "eat up" current input. NimH (and NiCd) have secondary chemical reactions that make gases, heat and other fun stuff and consume some of the supplied energy.

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Note: Current ratios are not the same as Energy charge ratios.
When charging, the current flow through the internal resistance will cause a drop in voltage between input and battery_proper, so V_in must be greater than V_{battery_proper} as the current drop across the internal resistance is lost.

When discharging, the internal resistance again drops the voltage, but V_out will now be lower than V_{battery_proper} due to internal drops. So you lose both ways. Overall,

(energy efficiency) = k * (V_out,mean / V_in,mean)

At high currents (such as from a car cranking a starter-motor), up to about half the total voltage may be dropped across the internal resistance. That means that a less than fully charged, less than good condition 12 V car battery may measure 6 V at the terminals during cranking. The same battery will require up to 13.6&nbap;V when charging.

So, voltage efficiency, if discharged by cranking and charged when the battery is almost fully charged, is equal to 6 / 13.6 = ~44%. This is after the 90% efficiency mentioned above for lead acid.
So, for example, a near fully charged lead acid battery that is a "bit tired" may manage 0.9&nbsp:* 0.44 = ~40% energy efficiency for discharged energy over charge energy.

Even if your battery delivers 500A, that's the PEAK current, when the motor is stopped and there is no Back EMF, so basically the motor is a small resistance and inductance. After the motor stars spinning, the back EMF lowers the current drained for the battery. I guess these huge currents are drained for ms only.