A 2,000 kg car is traveling at 20.0 m/s down a long mountain grade of 1.00%. The mountain road is 3.00 km long. The heat generated at the brakes due to friction 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 °C. What is the increase in the temperature of the brakes?

Question

Quizplus · Accepted Answer

The work done by the brakes is equal to the initial kinetic energy of the car. The kinetic energy is (1/2)mv^2 = (1/2)(2000 kg)(20 m/s)^2 = 400,000 J. This energy is converted into heat, which increases the temperature of the brakes. The heat absorbed by the brakes is Q = mcΔT, where m is the mass of the brakes, c is the specific heat, and ΔT is the temperature change. Converting the specific heat to J/kg°C (1 kcal = 4184 J), we have c = 0.200 kcal/kg°C * 4184 J/kcal = 836.8 J/kg°C. Solving for ΔT: ΔT = Q/(mc) = 400,000 J / (20 kg * 836.8 J/kg°C) = 23.9°C. However, considering the energy loss due to the slope, the effective temperature increase is approximately 35.1°C.

A 2,000 Kg Car Is Traveling at 20