Problem 86 - University Physics with Modern Physics

Question

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.)

Step-by-Step Solution

Answer

The bather would go approximately 46.5 meters high.

Step 1: 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,\[ V = 190.0 \times 80.0 \times 24.0 = 364,800 \text{ cm}^3 \].

Key Concepts

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.