Energy and Power problem (change in tempurature)

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David's car, weighing 1500 kg and traveling at 72 km/h, experiences a braking scenario where the brake pads, with a mass of 10 kg and a heat capacity of 300 J/kg°C, start at 20°C. The kinetic energy calculated is 300,000 J, which is primarily converted into heat energy during braking. The change in temperature of the brake pads is determined using the formula Q=mcΔT, resulting in a final temperature of 120°C. While some energy may be lost to sound and other factors, the calculation effectively shows the transformation of kinetic energy into heat. The conclusion reached is that the final temperature of the brake pads is 120°C.
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Homework Statement


David is traveling in his car (1500 kg) at 72 km/h. He sees a dog on the road and hits the brakes, what is the final temperature of his brake pads if they have a combined mass of 10 kg, a heat capacity of 300 J/kgC and they start at 20 degrees Celsius?


Homework Equations



Q=mcT
Change in T=T2-T1

The Attempt at a Solution



I am wondering if i have done this right

Ek=1/2mv^2
= 1/2*1500kg*20m/s^2
=300000
-3.0x10^5 J

change in temperature =Q/mc
=3.0x10^5 J/10 kg*300 J/kgC
=100 degrees C
100+20=120 degress Celsius

120 is final temperature of break
-Is this correct?
 
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Looks good! Some of the kinetic energy may be lost to sound, etc., but essentially, all of the kinetic energy is transformed to heat energy.
 
Thank you!
 

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