How much internal energy is produced in the battery

[abot]

I do not know what to solve for...?

One recargable battery of mass 15g delivers to a cd player an average current of 1.80mA at 1.60 V for 2.4hours before the battery needs to be recharged. The recharger maintains a potential difference of 2.30v across the battery and delivers a charging current of 13.5mA for 4.20 hours. (a) what is the efficiency of the battery as an energy storage device? (b) How much internal energy is produced in the battery during one charge - discharge cycle?

it sounds confusing. I tried solving for P, power but I am not sure that what they want. eficciency? I don't get this

Thanks in advanced

 

[Andrew Mason]

Double post

I do not know what to solve for...?

One recargable battery of mass 15g delivers to a cd player an average current of 1.80mA at 1.60 V for 2.4hours before the battery needs to be recharged. The recharger maintains a potential difference of 2.30v across the battery and delivers a charging current of 13.5mA for 4.20 hours. (a) what is the efficiency of the battery as an energy storage device? (b) How much internal energy is produced in the battery during one charge - discharge cycle?

it sounds confusing. I tried solving for P, power but I am not sure that what they want. eficciency? I don't get this

Thanks in advanced

This is also posted at:
https://www.physicsforums.com/showthread.php?t=69676

PF frowns on double posting.

AM

 

[ShawnD]

I do not know what to solve for...?

One recargable battery of mass 15g delivers to a cd player an average current of 1.80mA at 1.60 V for 2.4hours before the battery needs to be recharged. The recharger maintains a potential difference of 2.30v across the battery and delivers a charging current of 13.5mA for 4.20 hours. (a) what is the efficiency of the battery as an energy storage device? (b) How much internal energy is produced in the battery during one charge - discharge cycle?

As Andy said in the other thread of yours, efficiency is the amount in divided by the amount out.

amount out
E = IVt
E = (0.0018A)(1.6V)(2.4 hours)(3600 seconds/hour)
E = 24.8832 J

amount in
E = IVt
E = (0.0135A)(2.3V)(4.2 hours)(3600 seconds/hour)
E = 469.476 J

Divide the amount of power the battery provides by the amount of power put into the battery then multiply by 100 to get the efficiency in percent.

As for internal energy produced, I'm not exactly sure what that means. It's either asking for the amount of energy the battery releases, or the amount of energy wasted (energy in - energy out). I would guess it's for the amount of wasted energy.