A car of mass 1.25 103 kg is initially moving on a level road at a speed of 17.0 m/s. Compute the increase in temperature of the brakes, assuming that all the mechanical energy ends up as internal energy in the brake system. Assume a total heat capacity of 10,000 J/C°.
KE = m•v^2/2 = Q = C•Δt ,
Δt = m•v^2/2•C = 1250•(17)^2/2•10000 =18 degr
To compute the increase in temperature of the brakes, we need to calculate the mechanical energy of the car and then convert it into an increase in temperature.
Step 1: Calculate the initial kinetic energy of the car.
The kinetic energy (KE) of an object is given by the equation KE = 0.5 * mass * velocity^2.
Given:
Mass of the car (m) = 1.25 * 10^3 kg
Initial velocity (v) = 17.0 m/s
Using the formula, we can calculate the initial kinetic energy of the car:
KE = 0.5 * m * v^2
KE = 0.5 * 1.25 * 10^3 kg * (17.0 m/s)^2
KE ≈ 180,125 Joules
Step 2: Calculate the increase in temperature.
The mechanical energy gets converted into an increase in temperature of the brakes. We can use the formula:
ΔT = ΔQ / (m * C)
Where:
ΔT is the change in temperature (increase in temperature),
ΔQ is the change in heat energy,
m is the mass of the brake system, and
C is the heat capacity of the brake system.
Given:
C = 10,000 J/°C
Since all the mechanical energy (KE) is converted into heat energy (ΔQ), we can say that ΔQ = KE.
ΔT = KE / (m * C)
Substituting the values, we get:
ΔT = 180,125 J / (m * C)
Note: The mass of the brake system is not given in the question. We'll assume an average mass for the sake of calculation. Let's say the mass of the brake system is 10 kg.
ΔT = 180,125 J / (10 kg * 10,000 J/°C)
ΔT = 180,125 J / 100,000 J/°C
ΔT ≈ 1.80125 °C
Therefore, the increase in temperature of the brakes is approximately 1.80125 °C.