How much does a car's disk breaks heat up when,

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SUMMARY

The discussion focuses on calculating the heat generated in a car's disk brakes when decelerating from 120 km/h to 80 km/h. It establishes that 50% of the kinetic energy lost during deceleration translates into heat absorbed by the brakes. Using the formula Q = cmΔT, where Q represents heat energy, c is the heat capacity (0.46 for steel), m is the mass of the disk brakes (11 kg), and ΔT is the temperature change, participants aim to determine the temperature increase in the brakes due to this energy conversion.

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  • Understanding of kinetic energy and its conversion to heat energy
  • Familiarity with the formula Q = cmΔT for heat transfer calculations
  • Basic knowledge of thermodynamics, particularly heat capacity
  • Concept of deceleration and its impact on energy loss in vehicles
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  • Research the principles of kinetic energy and its formulas
  • Study the specific heat capacity of various materials, focusing on metals
  • Explore advanced thermodynamic equations related to heat transfer in automotive systems
  • Investigate the effects of brake material composition on heat dissipation
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Automotive engineers, physics students, and anyone interested in vehicle dynamics and thermal management in braking systems.

Northener
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Imaging a car decelerating from 120km/h to 80km/h. How much do the disk breaks heat up if 50% of the energy translates directly as heat (to the breaks). The car's mass is 1300kg and the full mas of the disk breaks are 11kg. The disks are made of steel Q=0.46 (heat capacity)

I've been using Ke formula and Q=cm¤T

¤ stands for ''difference in''. Not sure what the correct notation is.
 
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so you use the ke formula so find out the energy lost, you know that 50% of it translates to heat.

the other equation:
Q = cm \Delta T

Q is the heat energy that went into the break, c is the heat capacity, m is the mass of the disk, and \Delta T is the change in temperature of the break
 

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