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I have been given the following statistics for a battery and I have been asked to find the 1) battery capacity in Wh 2) volumetric energy density in Wh/cm3 3) gravimetric energy density in Wh/kg.

Volts: 3.7 V

Weight: 23 g

Capacity:1150 mAh

Dimensions: 55 x 34 x 5.5 mm

Could someone please check I am doing this right before i commit it to memory.

1) watt-hours = milliampere-hours * volts /1000

Watt-hours = 1150 mAh * 3.7 V / 1000

Answer = 4.255 Wh

2) To find the volumetric density in Wh/cm^3 so we use the formula Wh/cm^3

4.255Wh/10.29 cm ^3

Answer = 0.41cm ^ 3


To find the gravimetric energy density in Wh/Kg, we divide Wh by the weight of the battery in Kg (23g = 0.023Kg), so we have

4.255Wh / 0.023Kg = 185 Wh/Kg

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Looks ok to me. Do you have a specific question? What do you think you might be doing wrong? – Mark Beadles Feb 25 '12 at 15:34
Sorry forgot to add the stats for the battery, just unsure if i have got the conversions correct for the formulas, now added thanks – user445714 Feb 25 '12 at 15:39
This is an ill-posed problem, since you don't say what type of battery you have. What this means is that you pulled the energy per unit volume out of thin air, and the rest is mechanical calculations that are not worth someone else checking for you. I vote to close. – Ron Maimon Feb 25 '12 at 15:48
C-S2: Standard battery for the Blackberry Curve 8520, dont honestly know why that matters, i gave the battery statistics but anyhows – user445714 Feb 25 '12 at 15:53
Welcome to Physics.SE. Our FAQ does not look kindly on this kind of "solve my homeowrk problem" question, but you may find some help in Relating milliampere-hours to watt-hours for batteries. – dmckee Feb 25 '12 at 16:40

closed as too localized by Qmechanic, Ron Maimon, dmckee Feb 25 '12 at 16:38


1 Answer

Watch your units. For example, your answer to #2 is in $cm^3$ which is not a unit of volumetric energy density.

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