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An automobile has a mass of 1500 kg, and its aluminum brakes have an overall mass of 6.00 kg. (a) Assume all the mechanical energy that transforms into internal energy when the car stops is deposited in the brakes and no energy is transferred out of the brakes by heat. The brakes are originally at 20.0°C. What will be the temperature of the brakes when the car is stopped from 30.0 m/s.

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Final answer:

The temperature of the car's aluminum brakes will increase by 125°C when the car is stopped from 30.0 m/s, resulting in a final temperature of 145°C.

Step-by-step explanation:

The question relates to the conversion of kinetic energy into internal energy in the form of heat in a car's braking system and subsequently into an increase in the temperature of the brakes.

To find the temperature change of the brakes, we need to use the relationship between energy, mass, specific heat capacity, and temperature change (Q = mcΔT), where Q is the heat energy, m is the mass of the brakes, c is the specific heat capacity, and ΔT is the temperature change.

The kinetic energy of the car, which is all assumed to transform into heat within the brakes, is given by KE = 1/2mv², where m is the mass of the car and v is its initial velocity. As all this energy is used to heat the brakes, Q is equal to this kinetic energy. Given that the specific heat capacity of aluminum (the brake material) is approximately 900 J/kg·°C, we can use the formula mcΔT = 1/2mv² to find the increase in temperature ΔT.

We start by calculating the kinetic energy of the car:
KE = 1/2 × 1500 kg × (30.0 m/s)² = 675000 J

Then, we calculate the increase in temperature of the brakes:
mcΔT = KEΔT = KE / (mc)ΔT = 675000 J / (6.00 kg × 900 J/kg·°C)ΔT = 125 °C

The temperature of the brakes after stopping the car would then be 20.0°C + 125°C = 145°C.

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