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CP A person of mass 70.0 kg is sitting in the bathtub. The bathtub is 190.0 cm by 80.0 cm; before the person got in, the water was 24.0 cm deep. The water is at 37.0\(^\circ\)C. Suppose that the water were to cool down spontaneously to form ice at 0.0\(^\circ\)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.)

Short Answer

Expert verified
The bather would go approximately 46.5 meters high.

Step by step solution

01

Calculate the Volume of Water Displaced

The person sitting in the bathtub displaces a certain volume of water. The volume of water displaced is equal to the initial volume of water in the bathtub. The dimensions of the bathtub are 190.0 cm by 80.0 cm, and the initial water depth is 24.0 cm. To find the volume of the water displaced, we use the formula\[ V = L \times W \times D \]where \( L = 190.0 \) cm, \( W = 80.0 \) cm, and \( D = 24.0 \) cm. Thus,\[ V = 190.0 \times 80.0 \times 24.0 = 364,800 \text{ cm}^3 \].

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Key Concepts

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

Volume Displacement
Volume displacement is a key concept in physics, often encountered in problems involving fluids. When an object is submerged in a fluid, it pushes the fluid out of the way. The volume of fluid displaced is equal to the volume of the submerged part of the object. In our bathtub scenario, when a person sits in the tub, they displace a certain amount of water. This is crucial for calculating various effects like buoyancy and in our case, the potential to use energy due to phase changes.
To compute the volume of water displaced, we use a simple formula involving multiplication of the dimensions of the water section: the length, width, and depth. For the bathtub:
  • Length (L): 190.0 cm
  • Width (W): 80.0 cm
  • Depth (D): 24.0 cm
Make sure all measurements are in the same unit (here they are in centimeters). Multiply these together:\[ V = L \times W \times D = 190.0 \times 80.0 \times 24.0 = 364,800 \text{ cm}^3 \]
This gives us the total volume of the water displaced in the bathtub by the bather's mass.
Conservation of Energy
The conservation of energy principle states that energy cannot be created or destroyed, only transformed from one form to another. This is critical in analyzing many physical systems, particularly in thermodynamics.
In the context of our problem, when water in the bathtub cools from 37°C to 0°C and freezes into ice, it releases energy. According to the conservation of energy, this released energy can theoretically be redirected to do work, such as launching the person upwards.
Energy transformations in such processes can be calculated using the equation:\[ Q = m \times c \times \Delta T \]where:
  • \( Q \) is the heat energy transferred
  • \( m \) is the mass of the water
  • \( c \) is the specific heat capacity of water
  • \( \Delta T \) is the change in temperature
Furthermore, during the phase change from water to ice, the latent heat of fusion must be included. This calculation helps us understand the potential energy available to launch the bather into the air.
Thermodynamics
Thermodynamics refers to the branch of physics that deals with heat and temperature and their relation to energy and work. In our exercise, thermodynamics ties into the transformation processes that occur as the hot water in the bathtub cools and freezes.
According to the laws of thermodynamics, while energy transfer is feasible, the spontaneous cooling of water into ice and the resulting energy release conflicts with entropy principles. The system's entropy increases during spontaneous processes like cooling. However, using all released energy without any loss, as our scenario suggests, violates the second law of thermodynamics, which asserts that some energy scatters in less useful forms in real-life processes.
This is a good reminder that while calculations might suggest possibilities, natural laws, such as thermodynamics, govern the limitations of these processes. Energy transformations, entropy, and irreversible processes are fundamental topics under thermodynamics that help us properly analyze energy use and conversion in physical systems.

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Most popular questions from this chapter

Rods of copper, brass, and steel-each with crosssectional area of 2.00 cm\(^2\)-are welded together to form a Y-shaped figure. The free end of the copper rod is maintained at 100.0\(^\circ\)C, and the free ends of the brass and steel rods at 0.0\(^\circ\)C. Assume that there is no heat loss from the surfaces of the rods. The lengths of the rods are: copper, 13.0 cm; brass, 18.0 cm; steel, 24.0 cm. What is (a) the temperature of the junction point; (b) the heat current in each of the three rods?

Like the Kelvin scale, the Rankine scale is an absolute temperature scale: Absolute zero is zero degrees Rankine (0\(^\circ\)R). However, the units of this scale are the same size as those of the Fahrenheit scale rather than the Celsius scale. What is the numerical value of the triple-point temperature of water on the Rankine scale?

A 500.0-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 kg of water at room temperature (20.0\(^\circ\)C). After waiting and gently stirring for 5.00 minutes, you observe that the water’s temperature has reached a constant value of 22.0\(^\circ\)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) 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.

Styrofoam bucket of negligible mass contains 1.75 kg of water and 0.450 kg of ice. More ice, from a refrigerator at -15.0\(^\circ\)C, is added to the mixture in the bucket, and when thermal equilibrium has been reached, the total mass of ice in the bucket is 0.884 kg. Assuming no heat exchange with the surroundings, what mass of ice was added?

The African bombardier beetle (Stenaptinus insignis) can emit a jet of defensive spray from the movable tip of its abdomen (Fig. P17.91). The beetle's body has reservoirs containing two chemicals; when the beetle is disturbed, these chemicals combine in a reaction chamber, producing a compound that is warmed from 20\(^\circ\)C to 100\(^\circ\)C by the heat of reaction. The high pressure produced allows the compound to be sprayed out at speeds up to 19 m/s 168 km/h2, scaring away predators of all kinds. (The beetle shown in Fig. P17.91 is 2 cm long.) Calculate the heat of reaction of the two chemicals (in J/kg). Assume that the specific heat of the chemicals and of the spray is the same as that of water, \(4.19 \times 10{^3} J/kg \cdot K\), and that the initial temperature of the chemicals is 20\(^\circ\)C.

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