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The highest recorded waterfall in the world is found at Angel Falls in Venezuela. Its longest single waterfall has a height of \(807 \mathrm{~m}\). If water at the top of the falls is at \(15.0^{\circ} \mathrm{C}\), what is the maximum temperature of the water at the bottom of the falls? Assume all the kinetic energy of the water as it reaches the bottom goes into raising the water's temperature.

Short Answer

Expert verified
The maximum temperature of the water at the bottom of the falls is \( 16.89^{\circ}C \).

Step by step solution

01

Find the Change in Kinetic Energy

First, calculate the potential energy at the top of the waterfall that turns into kinetic energy at the bottom. The formula to calculate potential energy is \( PE = mgh \) where \( m \) is mass, \( g \) is acceleration due to gravity, and \( h \) is height. Since we don't have the mass, we use a unit mass of 1 kg, and gravity as \( 9.8 m/s^2 \). The change in kinetic energy, therefore, is \( dKE = g \times h = 9.8 \times 807 = 7908.6 J/kg \).
02

Calculate the Change in Temperature

The gain in kinetic energy is transformed into heat, hence raising the temperature of the water. The specific heat of water is 4184 J/kg•°C. The temperature change can be obtained from the formula \( \Delta T = \frac{dKE}{c} \) where \( c \) is the specific heat. Substituting the values, we get \( \Delta T = \frac{7908.6}{4184} = 1.89^{\circ}C \).
03

Find the Final Temperature

The original temperature of the water was given as \( 15.0^{\circ}C \). Adding the change in temperature to this, we get the final temperature of the water \( T = T_0 + \Delta T = 15 + 1.89 = 16.89^{\circ}C \).

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Key Concepts

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

Potential Energy
Potential energy is a form of energy that is stored due to the position of an object. For waterfalls, this energy is captured by the water at the top due to its elevated position. The higher the water, the greater the potential energy (PE). This energy can be calculated by using the formula \( PE = mgh \), where:
  • \( m \) is the mass of the water,
  • \( g \) is the acceleration due to gravity \( (9.8 \, \text{m/s}^2 )\),
  • \( h \) is the height of the waterfall.
Even though we don't always know the mass in these calculations and use a unit mass for simplicity, the height and gravity are crucial in determining how much energy can be stored as potential energy at the top of a waterfall. Once the water starts to fall, this potential energy begins to convert into kinetic energy.
Kinetic Energy
Kinetic energy takes over as water plunges over a waterfall and falls to the bottom. This type of energy depends on the motion or speed of the object, in this case, the water. Unlike potential energy, which is stored, kinetic energy is active energy that manifests when the water is moving. The conversion from potential to kinetic energy takes place as the water moves downward, guided by gravity. For water falling from a great height like Angel Falls, the initial potential energy calculated from the height becomes the total kinetic energy at the base, provided no energy is lost to the surroundings. We calculate the kinetic energy using a similar approach to potential energy and hence, all the gain is transferred to raising the water's temperature due to this motion.
Specific Heat Capacity
Specific heat capacity is crucial when it comes to understanding how substances like water absorb and hold heat. Defined as the amount of energy required to raise the temperature of 1 kilogram of a substance by 1 degree Celsius, it helps us determine how much temperature change can occur when energy is added. For water, this value is quite high, at 4184 J/kg°C. This high specific heat means water can absorb a lot of heat energy without a massive change in temperature, which is why it takes significant energy to increase the temperature, even slightly. In the case of the waterfall, this specific heat capacity defines how much the temperature of the water will rise once the kinetic energy is converted into heat.
Waterfalls
Waterfalls, like Angel Falls, beautifully demonstrate the principles of energy conversion. As one of nature's most awe-inspiring phenomena, they serve as excellent examples of how potential energy can be converted into kinetic energy, and eventually into heat energy. From a physics perspective, waterfalls allow us to witness energy in conversion, as the mass of water transitions from an elevated state to rushing downwards, transforming stored potential energy into energetic kinetic energy. Not only do they allow us to study these energy transformations, but they also highlight the efficiency and scale of natural processes, acting as both a spectacle and a natural laboratory to observe thermodynamics in action.
Temperature Change
The change in temperature as a result of energy conversion is an essential component of this scenario. It shows us how energy, previously potential and then kinetic due to the high drop, transforms into thermal energy - the heat which causes the water's temperature to rise.By calculating the energy involved and using the specific heat capacity, we can determine this temperature change. The formula for this transformation is \( \Delta T = \frac{dKE}{c} \), illustrating the direct relationship between kinetic energy and temperature increase. With waterfalls, when potential energy at the top is converted completely into kinetic energy as it reaches the base, this energy is then wholly used to increase the temperature of the water upon impact.

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Most popular questions from this chapter

\(\mid \mathbf{C}\) A \(3.00-g\) lead bullet at \(30.0^{\circ} \mathrm{C}\) is fired at a speed of \(2.40 \times 10^{2} \mathrm{~m} / \mathrm{s}\) into a large, fixed block of ice at \(0{ }^{\circ} \mathrm{C}\), in which it becomes embedded. (a) Describe the energy transformations that occur as the bullet is cooled. What is the final temperature of the bullet? (b) What quantity of ice melts?

Into a \(0.500-\mathrm{kg}\) aluminum container at \(20.0^{\circ} \mathrm{C}\) is placed \(6.00 \mathrm{~kg}\) of ethyl alcohol at \(30.0^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~kg}\) ice at \(-10.0^{\circ} \mathrm{C}\). Assume the system is insulated from its environment. (a) Identify all five thermal energy transfers that occur as the system goes to a final equilibrium temperature \(T\). Use the form "substance at \(X^{\circ} \mathrm{C}\) to substance at \(Y^{\circ} \mathrm{C} . "\) (b) Construct a table similar to the table in Example 11.5. (c) Sum all terms in the right-most column of the table and set the sum equal to zero. (d) Substitute information from the table into the equation found in part (c) and solve for the final equilibrium temperature, \(T\).

The temperature of a silver bar rises by \(10.0^{\circ} \mathrm{C}\) when it absorbs \(1.23 \mathrm{~kJ}\) of energy by heat. The mass of the bar is \(525 \mathrm{~g}\). Determine the specific heat of silver from these data.

A sprinter of mass \(m\) accelerates uniformly from rest to velocity \(v\) in \(t\) seconds. (a) Write a symbolic expression for the instantaneous mechanical power \(P\) required by the sprinter in terms of force \(F\) and velocity v. (b) Use Newton's second law and a kinematics equation for the velocity at any time to obtain an expression for the instantaneous power in terms of \(m, a\), and tonly (c) If a \(75.0-\mathrm{kg}\) sprinter reaches a speed of and \(t\) only. (c) If a \(75.0-\mathrm{kg}\) sprinter reaches a speed of \(11.0 \mathrm{~m} / \mathrm{s}\) in \(5.00 \mathrm{~s}\), calculate the sprinter's acceleration, \(11.0 \mathrm{~m} / \mathrm{s}\) in \(5.00 \mathrm{~s}\), calculate the sprinter's acceleration, assuming it to be constant. (d) Calculate the \(75.0-\mathrm{kg}\) sprinter's instantaneous mechanical power as a function of time \(t\) and (e) give the maximum rate at which he burns Calories during the sprint, assuming \(25 \%\) efficiency of conversion form food energy to mechanical energy.

A solar sail is made of aluminized Mylar having an emissivity of \(0.03\) and reflecting \(97 \%\) of the light that falls on it. Suppose a sail with area \(1.00 \mathrm{~km}^{2}\) is oriented so that sunlight falls perpendicular to its surface with an intensity of \(1.40 \times 10^{3} \mathrm{~W} / \mathrm{m}^{2}\). To what temperature will it warm before it emits as much energy (from both sides) by radiation as it absorbs on the sunny side? Assume the sail is so thin that the temperature is uniform and no energy is emitted from the edges. Take the environment temperature to be \(0 \mathrm{~K}\).

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