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Chapter 17: Problem 30

A 25,000-kg subway train initially traveling at 15.5 m/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 m long by 20.0 m wide by 12.0 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 kg/m\(^3\) and its specific heat to be 1020 J /kg \(\cdot\) K.

Short Answer

Expert verified
The air temperature in the station rises by approximately 0.156 K.

Step by step solution

01

Calculate Kinetic Energy of the Train

First, calculate the initial kinetic energy of the subway train using the formula for kinetic energy: \( KE = \frac{1}{2}mv^2 \), where \( m = 25000 \) kg is the mass of the train and \( v = 15.5 \) m/s is its initial velocity. \[ KE = \frac{1}{2} \times 25000 \times (15.5)^2 = 3009375 \text{ J} \]
02

Calculate Volume of the Station

Calculate the volume of the station using the formula for volume \( V = l \times w \times h \), where \( l = 65.0 \) m, \( w = 20.0 \) m, and \( h = 12.0 \) m. \[ V = 65.0 \times 20.0 \times 12.0 = 15600 \text{ m}^3 \]
03

Calculate Mass of Air in the Station

Find the total mass of the air inside the station using its volume and density. Use the formula \( m = \rho \times V \), where \( \rho = 1.20 \) kg/m\(^3\) is the density of air. \[ m = 1.20 \times 15600 = 18720 \text{ kg} \]
04

Find Temperature Change using Heat Transfer Formula

Use the formula for heat transfer \( Q = mc\Delta T \), rearranged to solve for \( \Delta T \) (change in temperature). Here, \( Q = 3009375 \) J is the heat transferred, \( m = 18720 \) kg is the mass of air, and \( c = 1020 \) J/kg\( \cdot\)K is the specific heat capacity of air. \[ \Delta T = \frac{Q}{mc} = \frac{3009375}{18720 \times 1020} \approx 0.156 \text{ K} \]
05

Interpret the Result

The calculated temperature change \( \Delta T \approx 0.156 \text{ K} \) indicates how much the air temperature in the station rises due to the heat release from the train's kinetic energy being converted to thermal energy.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy possessed by a moving object due to its motion. It鈥檚 a key concept in thermodynamics, particularly in understanding how moving objects can transfer energy. The formula for kinetic energy is given by:
  • \( KE = \frac{1}{2}mv^2 \)
where:
  • \( m \) is the mass of the object in kilograms (kg), and
  • \( v \) is the velocity of the object in meters per second (m/s).
In this context, the kinetic energy of the subway train is determined by its mass and velocity as it approaches the station. The initial kinetic energy is then released as heat when it comes to a stop.
Heat Transfer
Heat transfer is the process by which thermal energy moves from one body or system to another. In thermodynamic exercises like this, understanding heat transfer helps us interpret how energy changes form and affects surroundings. The principle here is that when the subway train stops, its kinetic energy is not lost but transferred as heat to the surrounding air in the station. Therefore, all the energy goes into warming up the air, demonstrating the conservation of energy where no energy is created or destroyed, only changed in form.
Specific Heat Capacity
Specific heat capacity is a property that tells us how much heat energy is needed to change the temperature of a given mass of a substance by one degree Celsius (or one Kelvin). The specific heat capacity of air in this example is 1020 J/kg路K, meaning that 1020 joules of energy are required to raise 1 kilogram of air by 1 degree Kelvin. This value is crucial when calculating the eventual temperature change experienced by the air in the station as a result of heat transfer from the train's kinetic energy.
Temperature Change
To calculate how the temperature of the air changes, we use the heat transfer equation:
  • \( Q = mc\Delta T \)
where:
  • \( Q \) is the heat transferred (in joules),
  • \( m \) is the mass of the object being heated (in kilograms),
  • \( c \) is the specific heat capacity, and
  • \( \Delta T \) is the change in temperature.
By rearranging to solve for \( \Delta T \), we find the temperature increase of the air inside the station, emphasizing how energy conversion directly affects temperature change. This process highlights the relationship between energy form conversion and thermal dynamics.
Mass and Density Calculations
Understanding mass and density is essential in comprehending how much matter is present and how tightly it is packed.To find the mass of the air in the station, we use:
  • Density formula: \( \rho = \frac{m}{V} \)
  • Rearranging gives us \( m = \rho \times V \)
For the given air density of 1.20 kg/m\(^3\) and volume of 15600 m\(^3\), the total mass of air can be calculated. This mass reflects how much air needs to be heated and plays a pivotal role in determining the temperature change when energy is transferred. Hence, density and mass provide foundational support for energy calculations in thermal physics.

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Most popular questions from this chapter

A U.S. penny has a diameter of 1.9000 cm at 20.0\(^\circ\)C. The coin is made of a metal alloy (mostly zinc) for which the coefficient of linear expansion is \(2.6 \times 10{^-}{^5} K{^-}{^1}\). What would its diameter be on a hot day in Death Valley (48.0\(^\circ\)C)? On a cold night in the mountains of Greenland (-53\(^\circ\)C)?

Convert the following Celsius temperatures to Fahrenheit: (a) -62.8\(^\circ\)C, the lowest temperature ever recorded in North America (February 3, 1947, Snag, Yukon); (b) 56.7\(^\circ\)C, the highest temperature ever recorded in the United States (July 10, 1913, Death Valley, California); (c) 31.1\(^\circ\)C, the world鈥檚 highest average annual temperature (Lugh Ferrandi, Somalia).

A metal rod that is 30.0 cm long expands by 0.0650 cm when its temperature is raised from 0.0\(^\circ\)C to 100.0\(^\circ\)C. A rod of a different metal and of the same length expands by 0.0350 cm for the same rise in temperature. A third rod, also 30.0 cm long, is made up of pieces of each of the above metals placed end to end and expands 0.0580 cm between 0.0\(^\circ\)C and 100.0\(^\circ\)C. Find the length of each portion of the composite rod.

Two rods, one made of brass and the other made of copper, are joined end to end. The length of the brass section is 0.300 m and the length of the copper section is 0.800 m. Each segment has cross-sectional area 0.00500 m\(^2\). The free end of the brass segment is in boiling water and the free end of the copper segment is in an ice鈥搘ater mixture, in both cases under normal atmospheric pressure. The sides of the rods are insulated so there is no heat loss to the surroundings. (a) What is the temperature of the point where the brass and copper segments are joined? (b) What mass of ice is melted in 5.00 min by the heat conducted by the composite rod?

You are given a sample of metal and asked to determine its specific heat. You weigh the sample and find that its weight is 28.4 N. You carefully add \(1.25 \times 10{^4}\) J of heat energy to the sample and find that its temperature rises 18.0 C\(^\circ\). What is the sample's specific heat?
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