This is from Sanjoy Mahajan's The Art of Insight Problem 1.11

Estimate the energy in a 9-volt battery. Is it enough to launch the battery into orbit?

We're just going to estimate the first part.

Battery Energy

A volt is energy per unit charge \( V = \frac{E}{q} \). To get towards an energy we need an amount of charge; the current in Ampere is the charge per unit time \( I = frac{q}{t} \). So the product \( V I = \frac{E}{t} \) is energy per unit time, or power.

My smoke detector needs a 9V battery, and should be replaced every year. If I can estimate the current the smoke detector draws I can estimate the energy.

The current drawn can be guessed with gut estimates based on my experience with current. 1 A is a lot of current, it's probably less than that, but it could be 100 mA. 1 mA is not much current it's probably more than that, but it could be 10 mA. Guessing the geometric mean of 1 A and 1 mA gives 30 mA, or 0.03 A.

Then the Energy in a 9V battery is given by

\( E = V I t = 9 \rm{V} \times 0.03 \rm{A} \times 3 \times 10^{7} s = 10^{7} \rm{J} \).

A common unit for energy in batteries is Watt Hours, which is 1/3600 J. So the energy is roughly 2500 Wh, or 2.5 kWh.

Checking

Wikipedia lists the typical capacity of 9V batteries as around 500 mAh. This then corresponds to around 5 Wh of energy. I've overestimated by a factor of 500.

It's easy to trace back where I went wrong; the most uncertain estimate was the current drawn by a smoke detector. According to energyrating.gov.au profile on Smoke Alarms the power drawn by a smoke alarm is less than 0.1 mW. So dividing the power by the 9 V gives a current drawn of around 0.01 mA.

It turns out my gut was completely wrong; I didn't really know enough about currents to make a gut estimate. I would have to think about another way to estimate the energy that doesn't rely on knowledge about currents that I don't have.