Gentlemen,
Good old ohm's law should apply. P (expressed in watts) = voltage x amps. Or, P, in this example is 600 watts, so solve the equation for amps (assume the voltage remains resonable constant in this case is 14.4 the volts [engine running], so the formula is P/E = I or 600 watts /14.4 volts = 41.6 amps, but this is an instantaneous value. One should consider the RMS value (average), or approximately 1/2 of the peak value. Will the alternator be capable? Likely yes, but one should pay attention to the supply wire size, perhaps a size #10 AWG. Capacitors can supply the additional instantaneous current needed for peak demand, which will be fsirly large in terms of capacitance, ecpressed as MFD or microfarads or even farads...again connecting wiring must be sized for the anticipated load. Actual total loads must be factored in to the requirment.....vehicial lights, blower motor, wiper motor, ignition, etc. The actual alternator selection must be by the user....there are many different brands and capacities available. Any questions, please ask or PM me.
Regards,
RJ Renton