Problem 24 - College Physics

Question

\(\bullet\) A \(25,000\) -kg subway train initially traveling at 15.5 \(\mathrm{m} / \mathrm{s}\) slows to a stop in a station and then stays there long enough for its brakes to cool. The station's dimensions are 65.0 \(\mathrm{m}\) long by 20.0 \(\mathrm{m}\) wide by 12.0 \(\mathrm{m}\) high. Assuming all the work done by the brakes in stopping the train is transferred as heat uniformly to all the air in the station, by how much does the air temperature in the station rise? Take the density of the air to be 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) and its specific heat to be 1020 \(\mathrm{J} /(\mathrm{kg} \cdot \mathrm{K})\)

Step-by-Step Solution

Answer

The air temperature in the station rises by approximately 0.155 K.

Step 1: Calculate Initial Kinetic Energy of the Train

Use the formula for kinetic energy: \[ KE = \frac{1}{2} m v^2 \]where \( m = 25,000 \, \text{kg} \) is the mass and \( v = 15.5 \, \text{m/s} \) is the initial velocity.\[ KE = \frac{1}{2} \times 25,000 \, \text{kg} \times (15.5 \, \text{m/s})^2 \]Calculating this gives:\[ KE = \frac{1}{2} \times 25,000 \times 240.25 \]\[ KE = 3,003,125 \, \text{J} \]

Step 2: Calculate the Volume of the Station

Find the volume of the station using its dimensions:\[ V = \text{Length} \times \text{Width} \times \text{Height} \]\[ V = 65.0 \, \text{m} \times 20.0 \, \text{m} \times 12.0 \, \text{m} \]\[ V = 15,600 \, \text{m}^3 \]

Step 3: Calculate the Mass of the Air inside the Station

Use the volume of the station and the density of air:\[ m_{\text{air}} = \text{density} \times \text{volume} \]where the density \( \rho = 1.20 \, \text{kg/m}^3 \).\[ m_{\text{air}} = 1.20 \, \text{kg/m}^3 \times 15,600 \, \text{m}^3 \]\[ m_{\text{air}} = 18,720 \, \text{kg} \]

Step 4: Calculate the Temperature Increase

Use the equation for heat transfer \( Q = mc\Delta T \), solving for \( \Delta T \):\[ Q = 3,003,125 \, \text{J} \] (from Step 1).\[ m = 18,720 \, \text{kg} \] (from Step 3).\[ c = 1020 \, \text{J/(kg} \cdot \text{K)} \].Rearrange the formula to find \( \Delta T \):\[ \Delta T = \frac{Q}{mc} \]\[ \Delta T = \frac{3,003,125}{18,720 \times 1020} \]\[ \Delta T \approx 0.155 \, \text{K} \]

Key Concepts

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. Imagine the subway train moving swiftly along its tracks – all that movement is kinetic energy at work.

To calculate kinetic energy, we use the formula:

• \[ KE = \frac{1}{2} m v^2 \]

where \( m \) is the mass and \( v \) is the velocity of the object.

For our subway train, with a mass of 25,000 kg and moving at a speed of 15.5 m/s, the initial kinetic energy turns out to be 3,003,125 Joules. This illustrates a fundamental principle: the faster or heavier something is, the more kinetic energy it stores.

Understanding kinetic energy is crucial in thermodynamics because there's often a trade-off between kinetic energy and heat energy, especially noticeable when objects slow down as their kinetic energy converts into other forms, like heat.

Heat Transfer

When the subway train comes to a stop, its kinetic energy doesn't just vanish. Instead, it transforms through heat transfer. This is a process where energy moves from one body or system to another in the form of heat.

In the case of the subway train, as it stops, the work done by the brakes converts the train's kinetic energy into thermal energy, which dissipates into the surroundings—in this instance, the air inside the station.

Heat transfer occurs through three primary methods:

• Conduction: Direct transfer of heat through a material.
• Convection: Heat transfer through fluid movement, like air or water.
• Radiation: Transfer of heat through electromagnetic waves.

In our scenario, the heat is primarily transferred through convection, as the heated air mixes within the station and raises the average temperature. Recognizing how heat moves is vital in thermodynamics, helping us predict temperature changes.

Specific Heat Capacity

Specific heat capacity is a property that tells us how much heat is needed to change an object's temperature by a certain amount. Different materials require different amounts of heat due to their specific heat capacities.

To find the temperature change (\( \Delta T \)) in the station, we employ the formula for heat transfer:

• \[ Q = mc\Delta T \]

where \( Q \) is the heat added or removed, \( m \) is the mass of the substance (in this case, air), and \( c \) is the specific heat capacity. For air, that specific heat capacity is 1020 J/(kg·K).

Plugging in our values, with the previously calculated mass of the air and the heat generated, we find the temperature increase in the station is approximately 0.155 K.

Understanding specific heat capacity is critical because it tells us just how responsive a material’s temperature might be to energy changes, which is a cornerstone in analyzing thermal systems.