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Chapter 3: Problem 123

A car with mass \(1275 \mathrm{~kg}\) is driven at \(60 \mathrm{~km} / \mathrm{h}\) when the brakes are applied quickly to decrease its speed to \(20 \mathrm{~km} / \mathrm{h}\). Assume that the brake pads have a \(0.5-\mathrm{kg}\) mass with a heat capacity of \(1.1 \mathrm{~kJ} / \mathrm{kg} \mathrm{K}\) and that the brake disks/drums are \(4.0 \mathrm{~kg}\) of steel. Further assume that both masses are heated uniformly. Find the temperature increase in the brake assembly.

Short Answer

Expert verified
The temperature increase in the brakes is approximately 66.15 K.

Step by step solution

01

Convert Speeds to m/s

Initially, the car travels at \( v_i = 60 \, \text{km/h} \). To convert this speed to \( \, \text{m/s} \, \), use the conversion factor: \( 1 \, \text{km/h} = \frac{1}{3.6} \, \text{m/s} \). Thus, \( v_i = 60 \times \frac{1}{3.6} = 16.67 \, \text{m/s} \).The final speed is \( v_f = 20 \, \text{km/h} \). Converting this gives \( v_f = 20 \times \frac{1}{3.6} = 5.56 \, \text{m/s} \).
02

Calculate Initial and Final Kinetic Energies

Calculate the initial kinetic energy: \( K_i = \frac{1}{2} \times m \times v_i^2 \), where \( m = 1275 \, \text{kg} \).\[ K_i = \frac{1}{2} \times 1275 \times (16.67)^2 \approx 177154.89 \, \text{J} \]Calculate the final kinetic energy: \( K_f = \frac{1}{2} \times m \times v_f^2 \).\[ K_f = \frac{1}{2} \times 1275 \times (5.56)^2 \approx 19673.28 \, \text{J} \]
03

Energy Dissipation in Brakes

The energy dissipated by the brakes is the difference between the initial and final kinetic energies: \( \Delta K = K_i - K_f \).\[ \Delta K = 177154.89 - 19673.28 \approx 157481.61 \, \text{J} \]
04

Calculate the Total Heat Capacity of the Brake Assembly

Calculate the total mass of the brake assembly: \( m_{\text{brakes}} = 0.5 + 4.0 = 4.5 \, \text{kg} \).Calculate the total heat capacity \( C \) (specific to mass):\[ C = 0.5 \times 1.1 + 4.0 \times 0.48 \approx 2.38 \, \text{kJ/K} \]Note: \( 1.1 \, \text{kJ/kg} \) for brake pads, \( 0.48 \, \text{kJ/kgK} \) is the specific heat for steel, expressed here intentionally.
05

Calculate Temperature Increase in Brake Assembly

Using the formula \( \Delta Q = C \cdot \Delta T \), where \( \Delta Q = 157481.61 \, \text{J} \) and \( C \approx 2380 \, \text{J/K} \), calculate:\[ \Delta T = \frac{\Delta Q}{C} = \frac{157481.61}{2380} \approx 66.15 \, \text{K} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy Calculation
Kinetic energy is the energy possessed by an object due to its motion. The formula is straightforward:
  • \( K = \frac{1}{2} \times m \times v^2 \)
Where:
  • \( K \) is the kinetic energy in joules (J)
  • \( m \) is the mass of the object in kilograms (kg)
  • \( v \) is the velocity of the object in meters per second (m/s)
In the exercise, the car's initial velocity is converted into m/s to ensure consistency with the units used in the kinetic energy formula.
This conversion is crucial as using inconsistent units can often lead to incorrect calculations.
After converting the speeds, we calculate both the initial and final kinetic energies with their respective velocities. This highlights how a change in speed affects kinetic energy.Understanding this step is key in determining the energy dissipation during the braking process.
Heat Capacity
Heat capacity reflects how much heat energy is needed to change the temperature of a given mass by 1掳C (or 1K). The total heat capacity of an object depends on both mass and material.
In the given problem, different components of the braking system, like brake pads and steel drums, have different heat capacities, specified in kJ/kgK.
  • For brake pads: \( 1.1 \, \text{kJ/kgK} \)
  • For steel: \( 0.48 \, \text{kJ/kgK} \)
The calculation combines the specific heat capacities of both materials, weighted by their respective masses, to find the overall heat capacity of the brake assembly.
This is crucial for understanding how the materials in the braking system respond to changes in temperature.
Energy Dissipation
Energy dissipation refers to the loss of energy as it transforms from one form to another, often into heat. During braking, kinetic energy converts into heat energy as the brakes slow down the car.
In the exercise, the difference between the initial and final kinetic energies of the car during braking is calculated to find out how much energy is dissipated.
  • Initial energy minus final energy gives us the amount dissipated
The brakes absorb this energy, and understanding this process helps explain how the temperature of the brake components increases.
It's important to recognize that though energy is "lost", it is actually conserved by transforming into heat which increases the temperature of brake parts.
Temperature Change Calculation
Calculating the temperature change of an object once energy is absorbed involves understanding both the total heat capacity and the energy dissipated.The formula used here is:
  • \( \Delta Q = C \cdot \Delta T \)
Where:
  • \( \Delta Q \) is the energy change (in joules)
  • \( C \) is the total heat capacity (in joules per kelvin, J/K)
  • \( \Delta T \) is the change in temperature (in kelvin, K)
By rearranging the formula to solve for \( \Delta T \), you divide the energy dissipated by the heat capacity. This calculation provides the temperature increase of the brakes.
Understanding this helps one appreciate the physical changes occurring in brake components during the braking process.

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Most popular questions from this chapter

A hydraulic hoist raises a 3650 -lbm car \(6 \mathrm{ft}\) in an auto repair shop. The hydraulic pump has a constant pressure of \(100 \mathrm{lbf} / \mathrm{in} .^{2}\) on its piston. What is the increase in potential energy of the car, and how much volume should the pump displace to deliver that amount of work?

Air in a piston/cylinder assembly at \(200 \mathrm{kPa}\) and \(600 \mathrm{~K}\) is expanded in a constant-pressure process to twice the initial volume, state \(2,\) as shown in Fig. P3.182. The piston is then locked with a pin, and heat is transferred to a final temperature of \(600 \mathrm{~K}\). Find \(P, T,\) and \(h\) for states 2 and \(3,\) and find the work and heat transfer in both processes.

A piston/cylinder assembly in a car contains \(0.2 \mathrm{~L}\) of air at \(90 \mathrm{kPa}\) and \(20^{\circ} \mathrm{C}\), as shown in Fig. \(\mathrm{P} 3.163 .\) The air is compressed in a quasi-equilibrium polytropic process with polytropic exponent \(n=1.25\) to a final volume six times smaller. Determine the final pressure and temperature, and the heat transfer for the process.

Air in a rigid tank is at \(100 \mathrm{kPa}, 300 \mathrm{~K}\) with a volume of \(0.75 \mathrm{~m}^{3}\). The tank is heated to \(400 \mathrm{~K}\), state 2 . Now one side of the tank acts as a piston, letting the air expand slowly at constant temperature to state 3 with a volume of \(1.5 \mathrm{~m}^{3} .\) Find the pressure at states 2 and 3 . Find the total work and total heat transfer.

Ammonia at \(0^{\circ} \mathrm{C}\) with a quality of \(60 \%\) is contained in a rigid 200 - \(\mathrm{L}\) tank. The tank and ammonia are now heated to a final pressure of 1 MPa. Determine the heat transfer for the process.
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