/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 A \(25,000-\mathrm{kg}\) subway ... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(25,000-\mathrm{kg}\) 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 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\).

Short Answer

Expert verified
The temperature rise is approximately 0.53 K.

Step by step solution

01

Calculate Initial Kinetic Energy of the Train

The initial kinetic energy (K.E.) of the train can be calculated using the formula: \[ K.E. = \frac{1}{2} m v^2 \]where \( m = 25000 \, \text{kg} \) is the mass of the train and \( v = 15.5 \, \text{m/s} \) is its initial velocity. Therefore, \[ K.E. = \frac{1}{2} \times 25000 \times (15.5)^2 \].

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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 a type of energy that a moving object possesses because of its motion. It is given by the formula \( K.E. = \frac{1}{2} m v^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. In our subway train example, the train possesses kinetic energy because it is moving at an initial velocity of 15.5 meters per second before it slows down to a stop. To calculate the initial kinetic energy of the train:- Plug in the mass of the train, \( m = 25000 \, \text{kg} \), and the velocity, \( v = 15.5 \, \text{m/s} \), into the equation.- The initial kinetic energy is found to be: \[ K.E. = \frac{1}{2} \times 25000 \times (15.5)^2 \]Understanding how kinetic energy works helps us predict how much work is needed to bring the train to a halt. In this case, all the work done to stop the train is converted into heat energy, which is why knowing how to calculate kinetic energy is essential for our problem.
Specific Heat Capacity
Specific heat capacity is a physical property that shows how much energy is needed to change the temperature of a unit mass of a substance by one degree Celsius. It is represented by the symbol \( c \) and measured in joules per kilogram per Kelvin (J/kgâ‹…K). In the context of the subway station scenario, we are particularly interested in the specific heat capacity of air because the heat generated from stopping the train is transferred to the air in the station. This property is crucial in determining how much the temperature of the air will rise. The given specific heat capacity of air is 1020 J/kgâ‹…K. This means for every kilogram of air, 1020 joules of energy is needed to increase its temperature by one Kelvin. Knowing the specific heat capacity allows us to calculate the temperature change by utilizing the relationship:- \( Q = mc\Delta T \), where \( Q \) is the heat absorbed, \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the temperature change. Utilizing specific heat capacity helps us figure out the relationship between the energy change and the temperature variation in this problem.
Heat Transfer
Heat transfer is the process by which heat energy is exchanged between physical systems. In our subway problem, heat transfer occurs when the kinetic energy of the moving train is converted into heat energy by the brakes. This heat is then distributed uniformly to the air in the subway station.The total heat energy transferred, \( Q \), equals the initial kinetic energy of the train. To find out how this affects the air temperature in the station, follow these steps:
  • First, calculate the volume of the station from its dimensions: 65.0 m long, 20.0 m wide, and 12.0 m high, yielding \(65.0 \times 20.0 \times 12.0 = 15600 \text{ m}^3\).
  • Using the density of air, \( \rho = 1.20 \text{ kg/m}^3\), calculate the mass of the air: \( m = 1.20 \times 15600\).
  • Apply the formula \( Q = mc\Delta T \) to find \( \Delta T \), where \( Q \) is the total heat from kinetic energy, and \( c \) is the specific heat capacity (1020 J/kgâ‹…K).
Understanding heat transfer in this context helps to determine quantitatively how the energy conversion affects the air temperature, which is the problem's requirement.

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Most popular questions from this chapter

A vessel whose walls are thermally insulated contains 2.40 \(\mathrm{kg}\) of water and 0.450 \(\mathrm{kg}\) of ice, all at a temperature of \(0.0^{\circ} \mathrm{C}\) . The outlet of a tube leading from a boiler in which water is boiling at atmospheric pressure is inserted into the water. How many grams of steam must condense inside the vessel (also at atmospheric pressure) to raise the temperature of the system to \(28.0^{\circ} \mathrm{C}\) ? You can ignore the heat transferred to the container.

A person of mass 70.0 \(\mathrm{kg}\) is sitting in the bathtub. The bathtub is 190.0 \(\mathrm{cm}\) by 80.0 \(\mathrm{cm}\) ; before the person got in, the water was 16.0 \(\mathrm{cm}\) deep. The water is at a temperature of \(37.0^{\circ} \mathrm{C}\) . Suppose that the water were to cool down spontaneously to form ice at \(0.0^{\circ} \mathrm{C},\) and that all the energy released was used to launch the hapless bather vertically into the air. How high would the bather go? (As you will see in Chapter 20 , this event is allowed by energy conservation but is prohibited by the second law of thermodynamics.)

A glass vial containing a 16.0 -g sample of an enzyme is cooled in an ice bath. The bath contains water and 0.120 \(\mathrm{kg}\) of ice. The sample has specific heat 2250 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) ; the glass vial has mass 6.00 \(\mathrm{g}\) and specific heat 2800 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K} .\) How much ice melts in cooling the enzyme sample from room temperature \(\left(19.5^{\circ} \mathrm{C}\right)\) to the temperature of the ice bath?

\(\mathrm{A} 500.0-\mathrm{g}\) chunk of an unknown metal, which has been in boiling water for several minutes, is quickly dropped into an insulating Styrofoam beaker containing 1.00 \(\mathrm{kg}\) of water at room temperature \(\left(20.0^{\circ} \mathrm{C}\right) .\) After waiting and gently stirring for 5.00 minutes, you observe that the water's temperature has reached a constant value of \(22.0^{\circ} \mathrm{C}\) (a) Assuming that the Styrofoam absorbs a negligibly small amount of heat and that no heat was lost to the surroundings, what is the specific heat of the metal? (b) Which is more useful for storing thermal energy: this metal or an equal weight of water? Explain. (c) What if the heat absorbed by the Styrofoam actually is not negligible. How would the specific heat you calculated in part (a) be in error? Would it be too large, too small, or still correct? Explain.

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 \(\mathrm{N}\) . You carefully add \(1.25 \times 10^{4} \mathrm{J}\) of heat energy to the sample and find that its temperature rises 18.0 \(\mathrm{C}^{\circ} .\) What is the sample's specific heat?

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