How Hot Will a Pack Get? Estimating Temperature Rise from Resistance and Thermal Mass

This is the third piece in a short series on heat. We have looked at how sustained heat ages a lithium pack, and how internal resistance both saps performance and creates heat. Here is the question that ties it together: for a given load, how much will a pack actually warm up? Answer that and you can design and run the pack to stay in the window where it performs well and lasts.

Two things decide the answer. One is how much heat the pack makes. The other is how much energy it takes to warm the pack up, a property called its thermal mass. We will start with the thermal mass, partly because it comes first in the logic and partly because measuring it is a satisfying little experiment.

Measuring the thermal mass, with ice

You do not have to take a thermal-mass figure on faith. It is something you can measure directly, and the way we pinned it down for a cell is a simple piece of calorimetry that anyone with a scale and a thermometer can follow.

The idea is to use melting ice as a known heat sink. We took a sample of ice and brought it right up to the melting point, to 0 degrees, so that any heat entering it would go entirely into melting it rather than warming it first. We weighed that ice, because the energy it takes to melt a given mass of it, roughly 334 joules per gram, is a well-known constant. Then we let a well-insulated cell melt the ice from the bottom, and as the water formed we wicked it away so it could not sit there and soak up heat of its own and skew the result. The only heat we wanted to count was the heat that went into the melting.

We measured the cell's temperature before and after. In giving up heat to melt the ice, the cell's own temperature dropped, and that drop is the key. The heat that left the cell has to equal the heat that melted the ice, so the cell's mass times its specific heat times the temperature it fell equals the mass of ice times the energy to melt ice:

cell mass × specific heat × temperature drop = ice mass × energy to melt ice

Everything there is known or measured except the specific heat, so rearrange to solve for it:

specific heat = (ice mass × energy to melt ice) ÷ (cell mass × temperature drop)

The mass of the cell sits right in the denominator, which makes sense: for the same drop in temperature, a heavier cell must have given up more heat, so it holds more energy per degree. The result is the cell's specific heat, the number we carry into the rest of the calculation.

The heat side: current, resistance, and time

The other ingredient is the heat the pack makes, and it comes from resistance. Every amp that flows through a pack's internal resistance makes heat. The rate at which heat appears is the current times itself, times the resistance. Multiply that by how long the current flows and you have the total heat produced:

Heat = current² × resistance × time

The important feature is that little squared. The heat does not follow the current, it follows the current multiplied by itself. Double the current and you do not double the heat, you quadruple it. This is why a pack can run cool while it loafs along and then warm up quickly the moment it is asked for a hard pull.

A quick illustration. Say a section of a pack has an internal resistance of 10 milliohms and is delivering 100 amps during a hard pull. The heat it makes is 100 × 100 × 0.010, which comes to 100 watts. Run that for one minute and you have put roughly 6,000 joules of heat into the pack. These numbers are illustrative, chosen to keep the arithmetic clean, not figures from any particular pack.

The temperature side: putting it together

Now both ingredients are in hand. The temperature rise is the heat the pack makes divided by its thermal mass, and thermal mass is the pack's mass times the specific heat we measured with the ice:

Temperature rise = (current² × resistance × time) ÷ (mass × specific heat)

Back to the example. Suppose that section holds about 2 kilograms of cells, and a lithium cell takes very roughly 1,000 joules to warm each kilogram by one degree, in the ballpark of what a measurement like the one above gives. That is a thermal mass of about 2,000 joules per degree. Divide the 6,000 joules of heat by 2,000 and you get a temperature rise of about 3 degrees over that minute.

Now watch what the square does. Keep everything the same but double the current to 200 amps. The heat rate does not double, it quadruples, to 400 watts. Over the same minute that is 24,000 joules, and the temperature rise jumps to about 12 degrees. Same pack, twice the current, four times the heating.

What the estimate does and does not tell you

This is a first-order, worst-case estimate, and it is worth being honest about that. It assumes every bit of heat stays inside the pack, with nothing escaping to the surrounding air or to a cooling system. Real packs shed heat, so the actual temperature rise is lower than the number this gives, and the gap grows the longer the load lasts. For a short burst the estimate is close. For sustained work, cooling is exactly what keeps the real number down. Treat this as a back-of-envelope tool for building intuition, not a specification or a safety prediction.

Even as a rough tool, though, it points straight at the levers that matter, and they are the same ones from the first two articles:

Because heat grows with the square of the current while the energy you actually move grows only in step with it, going slower makes less heat for the same work done. Push twice as hard and you deliver energy twice as fast but generate heat four times as fast. That is the arithmetic behind charging slower when you have the time, the point we made in the lifespan article about reaching a lower peak temperature.

More thermal mass rides through a burst with a smaller rise, so larger cells buy real thermal headroom. Lower resistance, from good cells and solid, low-resistance connections, means less heat at every amp, which is the whole argument of the resistance article. And when the duty cycle asks for more than mass and low resistance can absorb on their own, that is where active cooling earns its place.

Across the three articles the story is simple. Heat is what ages a pack, resistance is what makes the heat and steals performance, and thermal mass, with one short calculation, tells you how hot the pack will get. Design a pack that stays cool and you have designed one that lasts. If you want to work through the right approach for your equipment, our Battery Designer lets you explore configurations, or you can talk to an engineer and we will help you find the right fit.