Now that we have estimated the mass of the brakes, we can perform some analysis
Let's start with a case where the aircraft is rolling at 30kts, and we want to brake to a stop.
First we will find the energy in the moving aircraft: $$30\text{kts}=15\tfrac{m}{s}$$ $$KE = \frac{(2.7\times10^4kg)(15ms^{-1})^2}{2}$$ $$=3.04 \times 10^6 J$$
Solving the thermal energy equation for temperature: $$\Delta T=\frac{q}{mC}$$ $$=\frac{3.04 \times 10^6 J}{386\text{kg} \cdot 500Jkg^{-1}K^{-1}}$$ $$=16^\circ K (\text{or }C)$$ A brake application from 30kts to a complete stop raises the temperature 16°C.
Lets's consider now with engines at idle. At idle the engines are producing thrust. They are producing enough thrust to accelerate the aircraft. We like to taxi single engine to save fuel, but even with one engine at idle, a light CRJ700 will still accelerate. My operator does not permit the use of idle thrust reverse to null taxi thrust although most high mounted engine operators do.
The CRJ700 has two engines rated at a max thrust of 13,790 pounds of force which is equivalent to 61kN (kilo Newtons). The manufacturer does not publish idle thrust, so I have once again resorted to browsing the un-citeable web forums. Idle thrust being 5% of max thrust is what I am going to go with here.
To get energy (J) from a force, the thrust, we need to multiply force × distance. My best guess at how early I would start braking for a gentle stop from 30kts would be about 200 meters. That's 600 feet or two thirds the distance along a runway from the beginning to the aim point markers.
Therefore the brakes must absorb the energy from the mass of airplane moving at 30kts as calculated earlier in addition to the thrust of the engines during the braking event.
The work (energy) produced by 2 idling engines while braking over a 200m distance: $$\text{Work} = \text{Force} \cdot \text{Distance}$$ $$=(2\times 0.05\cdot61,000 N) \cdot 200m$$ $$=1.2\times10^6J$$
Adding this to our stopping energy from earlier: $$\Delta T=\frac{3.04 \times 10^6 J+1.2\times10^6J}{386\text{kg} \cdot 500Jkg^{-1}K^{-1}}$$ $$=22^\circ K (\text{or }C)$$ Running the math for a single engine taxi yields 19°C increase in brake temperatures.
Now 15-22°C may not sound like a lot, but just two brake applications is enough to increase the BTMS by one value.
The energy from the engines is dependent on distance too so stopping in a shorter distance would cause the increase in temperature to be closer to the 15°C value.
Due to the idle thrust of two and even just one engine, brakes must be used during taxi. There are two techniques, you can ride the brakes, or you can allow the aircraft to accelerate, then brake to a slower speed, and repeat the cycle.
Riding the brakes at a steady speed means the energy absorbed by the brakes is exactly the same as the energy output by the engine (ignoring other friction).
For aircraft with carbon brakes, the recommended taxi technique is to allow speed to build from 10kts to 30kts at idle, then brake back to 10kts and repeat. Aircraft Braking Systems Corp calls this technique 'snubbing' (source). Snubbing means the brakes absorb the kinetic energy of the aircraft slowing from 30 to 10kts, which must occur at periodic intervals during the taxi. (Carbon brakes wear less at higher temperatures).
Some captains at my operator are taught to use the snubbing technique and to avoid the brake riding technique. Why? I am not sure. Asking different captains yields unconfident answers, but something along the lines of riding the brakes causes them to be hotter. Is this true? Let's do some math and find out.
To recap, the change in temperature of the brakes can be expressed as: $$\Delta T=\frac{\Delta\text{KE}+(\text{force of 2 idling engines}\times\text{distance of braking event})}{\text{mass of wheel brakes}\times \text{specific heat of steel}}$$
Riding
Because riding the brakes is performed at a constant velocity, the change in kinetic energy is 0. The distance of the braking event is the distance of the taxi (excluding the beginning and end of taxi). $$\Delta T_\text{riding}=\frac{F_\text{engines}\cdot d_\text{taxi}}{m_\text{brakes}\cdot C}$$
Snubbing
For snubbing the change in kinetic energy is the change from our maximum taxi speed back down to our minimum taxi speed plus the force of engines over the braking event. The technique advises cycling between 30kts and 10kts, but I will leave the speeds as variables. $$\Delta T_\text{snubbing}=\frac{\frac{m_\text{aircraft}}{2}(v_\text{max}^2-v_\text{min}^2)+(F_\text{engines}\cdot d_\text{brake application})}{m_\text{brakes}\cdot C}$$
Comparison
Lets determine how much distance one of these cycles takes. The total distance will be distance for acceleration from the minimum speed to maximum speed, plus the distance for the braking event: $$d_\text{total}=d_\text{accel}+d_\text{brake application}$$
From \(F=m\cdot a\): $$a = \frac{F_\text{engines}}{m_\text{aircraft}}$$
Creating a function of velocity versus time with the aircraft starting at minimum taxi speed at time=0: $$v = \frac{F_\text{engines}}{m_\text{aircraft}}\cdot t + v_\text{min}$$
We will skip the integral math today, the distance traveled during acceleration is expressed as: $$d_{accel} = \frac{1}{2}\frac{F_\text{engines}}{m_\text{aircraft}}\cdot t_\text{max speed}^2 + v_\text{min} \cdot t_\text{max speed}$$
The amount of time it takes to reach maximum taxi speed: $$v_\text{max} = \frac{F_\text{engines}}{m_\text{aircraft}}\cdot t_\text{max speed} + v_\text{min}$$ $$t_\text{max speed}=(v_\text{max}-v_\text{min}) \frac{m_\text{aircraft}}{F_\text{engines}}$$
Putting it together to determine the distance: $$d_\text{accel} = \frac{1}{2}\frac{F_\text{engines}}{m_\text{aircraft}}\cdot \left((v_\text{max}-v_\text{min}) \frac{m_\text{aircraft}}{F_\text{engines}}\right)^2 + v_\text{min} \cdot \left((v_\text{max}-v_\text{min}) \frac{m_\text{aircraft}}{F_\text{engines}}\right)$$ which simplifies to: $$d_\text{accel}=\frac{m_\text{aircraft}}{2f_\text{eng}}(v_\text{max}^2-v_\text{min}^2)$$
Going back to our equation for riding the brakes, lets look at riding the brakes for the distance of one cycle of snubbing: $$\Delta T_\text{riding}=\frac{F_\text{engines}\cdot (d_\text{accel}+d_\text{brake application})}{m_\text{brakes}\cdot C}$$ $$=\frac{F_\text{engines}\cdot d_\text{accel}+F_\text{engines}\cdot d_\text{brake application}}{m_\text{brakes}\cdot C}$$ $$=\frac{F_\text{engines} \left(\frac{m_\text{aircraft}}{2f_\text{eng}}(v_\text{max}^2-v_\text{min}^2)\right) +F_\text{engines}\cdot d_\text{brake application}}{m_\text{brakes}\cdot C}$$ $$=\frac{\frac{m_\text{aircraft}}{2}(v_\text{max}^2-v_\text{min}^2) +F_\text{engines}\cdot d_\text{brake application}}{m_\text{brakes}\cdot C}$$ And this equation is exactly the same as our \( \Delta T_\text{snubbing} \) equation.
I did this in generic variables so it doesn't matter if my estimates of brake mass or idle thrust output are correct. What matters is the mathematical relationship of the variables involved, and it appears the brake riding and snubbing techniques incure identical penalties on brake temperature.