What if a battery wasted energy as heat?

Olly's battery holds energy. It runs a little lamp.

But some energy might turn into heat on the way to the lamp. What if a lot of it did?

A battery's label says how much energy it can store, in watt-hours. A device's label says how fast it uses energy, in watts. Divide one by the other, and you'd expect to know exactly how long the battery will last.

But that sum leaves something out: some energy is lost along the way, mostly as heat. In this lab, you make the trip from battery to lamp less and less efficient and see what happens to the runtime.

Energy is conserved in every conversion, but OpenStax notes that the useful output is always less than the input; the rest is transformed into other forms, such as thermal energy. Efficiency is useful output ÷ total input.

This lab uses a steady-load battery model: a 60 Wh battery, a 10 W lamp, all of the capacity usable, and a conversion efficiency you can lower from 100% to 70%. It asks how runtime and delivered energy change.

In a perfect world, all the battery's energy goes to the lamp.

The lamp shines for 6 whole hours.

In the model, the battery stores 60 watt-hours and the lamp uses 10 watts. With 100% efficiency, every bit of the stored energy reaches the lamp: 60 watt-hours.

So the lamp runs for 60 ÷ 10 = 6 hours. A watt-hour is the energy used by something drawing 1 watt for an hour, so a 10-watt lamp uses 10 watt-hours every hour.

With efficiency 100%, deliverable energy equals capacity: 60 Wh. Since E = P·t, the runtime is t = E ÷ P = 60 Wh ÷ 10 W = 6 h. OpenStax uses the same relation for household bills: energy used is power times time, measured in kilowatt-hours.

This ideal case is the label calculation: it assumes no losses anywhere between the cells and the device.

Now Olly lets some energy turn into heat. Only part of it reaches the lamp.

The lamp goes dark sooner, after about 4 hours.

Set the efficiency to 70%. Now only 70% of the battery's energy reaches the lamp: 42 watt-hours. The other 18 watt-hours turn into heat on the way.

The lamp still uses 10 watts, so it runs for 42 ÷ 10 = 4.2 hours, which is 4 hours 12 minutes. That's almost 2 hours less than the label sum. At 90% efficiency, the lamp would run 5 hours 24 minutes; at 80%, 4 hours 48 minutes.

At 70% efficiency the deliverable energy is 60 × 0.7 = 42 Wh, so t = 42 ÷ 10 = 4.2 h (4 h 12 min), and 18 Wh is dissipated as heat. Runtime is directly proportional to efficiency: 90% gives 5.4 h, 80% gives 4.8 h.

To recover 6 hours at 70% efficiency, capacity must rise to about 86 Wh (86 × 0.7 = 60.2 Wh, 6 h 1 min in the model). Losses therefore cost either runtime or a larger, heavier battery.

Energy never disappears. When it doesn't reach the lamp, it warms things up instead.

That warmth is energy the lamp never got to use.

Energy can't vanish. When it changes from one form to another, some always ends up as a form we didn't want, usually heat. OpenStax, a free physics textbook, explains that electric current heats the wires it flows through. In our model, the missing energy stands for heat in the wires, chargers and other electronics between the battery and the device.

Our model keeps it simple. It uses one efficiency number for every loss, keeps the lamp's power perfectly steady, and lets the battery give all its stored energy whatever the temperature, age or power. Real batteries do none of those things perfectly.

Conservation of energy means the 18 Wh is not destroyed: OpenStax describes efficiency as the fraction left as useful energy rather than transformed into, for example, thermal energy. In electrical systems that includes I²R heating in conductors and losses in converters.

Model limits, from its documentation: a single efficiency stands for all losses; the load is constant; the battery delivers its full usable capacity whatever the power, temperature or age (real batteries give less when cold, old or at very high power, as internal resistance turns more energy into heat); and only energy is tracked, not voltage or current.

Giant batteries that store electricity for the power grid lose some energy too. In 2019, they gave back about 8 out of every 10 parts they stored.

Electric cars lose a little energy when they charge.

Real systems lose energy too. The U.S. Energy Information Administration reported that in 2019, the country's big utility-scale batteries gave back about 82% of the electricity put into them. That's called round-trip efficiency.

Electric cars lose some energy while charging, says FuelEconomy.gov, when electricity is changed from the grid's alternating current to the direct current a battery uses, and in overcoming the battery's resistance to charging. How much is lost depends on the car, the charger, the battery's state and the weather.

The EIA defines round-trip efficiency as the percentage of electricity put into storage that is later retrieved. In 2019 the U.S. utility-scale battery fleet averaged a monthly round-trip efficiency of 82%, and pumped-storage hydropower 79% (EIA, 2021).

FuelEconomy.gov attributes electric-vehicle charging losses to converting AC from the grid to DC for the battery and to overcoming the battery's resistance to charging, which increases as the battery nears capacity; it puts the electric drive system's loss at 15% to 20%. OpenStax's efficiency table, for comparison, lists an electric motor at 98%.