A 2000 kg car is traveling at 20.0 m/s and then puts on the brakes and comes to a complete stop.The heat generated at the brakes due to stopping heats up the brakes.The mass of the brake system is 20.0 kg and it has a specific heat of 0.200 kcal/kg$^{\circ}$C.What is the increase in the temperature of the brakes?

Question

Quizplus · Accepted Answer

First, we need to find the total heat energy generated by the brakes:
KE = (1/2)mv^2 = (1/2)(2000 kg)(20.0 m/s)^2 = 400,000 J
The heat energy generated by the brakes is equal to the heat energy absorbed by the brake system:
Q = mcΔT
Where m is the mass of the brake system, c is the specific heat, and ΔT is the change in temperature.
Plugging in the given values, we get:
400,000 J = (20.0 kg)(0.200 kcal/kg°C)(ΔT)
Converting units:
400,000 J = (20.0 kg)(4186 J/kg°C)(ΔT)
ΔT = 1.90°C
Converting to Fahrenheit:
ΔT = 1.90°C x (9/5) + 32°F = 23.9°F
Therefore, the increase in temperature of the brakes is 23.9°F, which corresponds to choice C.

A 2000 Kg Car Is Traveling at 20 ∘^{\circ}∘ CWhat Is the Increase in the Temperature of the Brakes

A 2000 Kg Car Is Traveling at 20 $^{\circ}$ CWhat Is the Increase in the Temperature of the Brakes