This past week, I paid attention to brake temperatures following my landings. I usually began brake application around 90kts and was flying a 60,000lb aircraft.
I observed brake temperatures range from 03-04 within 15 minutes; temperatures of 100-150°C.
The forces involved in stopping were primarily the brakes, however there is also idle reverse thrust, drag, and wheel axle friction to at work. I am going to consider these effects negligible and assume that all the energy of motion was converted into heat in the brakes.
Kinetic energy can be expressed as:
$$\text{KE}=\frac{mV^2}{2}$$
where \(\text{KE}\) is kinetic energy in joules (\(J\)),
\(m\) is the mass in kilograms,
and \(V\) is the velocity in meters per second (\(ms^{-1}\)).
Converting to SI units: $$90\text{kts}=46.3ms^{-1}$$ $$60,000\text{lbs}=2.7\times10^4kg$$
Calculating the energy absorbed by the wheel brakes: $$\text{KE}=\frac{(2.7\times10^4kg)(46.3ms^{-1})^2}{2}=2.89\times10^7J$$
Heat energy can be expressed via the following relationship:
$$q=mC\Delta T$$
where \(q\) is thermal energy in joules,
\(m\) is the mass of the substance in kilograms,
\(C\) is the specific heat capacity of a material in units of joules per kilogram Kelvin (\(Jkg^{-1}K^{-1}\)),
and \(\Delta T\) is the change in temperature of a substance in units of Kelvin (\(K\)).
The wheel brakes are steel, and steel has a specific heat capacity of about 500 \(Jkg^{-1}K^{-1}\). (https://www.engineersedge.com/materials/specific_heat_capacity_of_metals_13259.htm)
Let's assume the brakes start at a temperature of 0°C and increase to a temperature of 150°C based on my observation of BTMS values of 3s and 4s. That's a change of 150° in both Celsius and Kelvin.
We now have all we need to take a stab at guessing the mass of the aircraft's brakes.
Solving the heat energy equation for mass: $$m=\frac{q}{C\Delta T}=\frac{2.89\times10^7J}{500Jkg^{-1}K^{-1}\times150^\circ K}=386\text{kg}$$ $$386\text{kg}=850\text{lbs}$$ $$\approx200\text{lbs per wheel}$$
Despite the crudeness of my anecdotal data collection, research on web forums leads me to believe that 200lbs per brake might actually be pretty close.
There is a very important point to be made here. The variables that go into brake temperature are simply mass and velocity.
We are ignoring the effects of drag and idle reverse thrust. Those forces would eventually stop the airplane, but it would be in excess of 12,000ft (take a look at the landing performance chart for an icy runway and no thrust reverse). The wheel brakes predominate.
It does not matter how long or how short your rollout is, it does not matter if you stand on the brakes or apply light pressure, the impact on brake temperature is almost the same. What matters is the aircraft weight, and the speed at which the brakes begin slowing the aircraft.