Problem 29 - University Physics with Modern Physics

Question

CP While painting the top of an antenna 225 m in height, a worker accidentally lets a 1.00-L water bottle fall from his lunchbox. The bottle lands in some bushes at ground level and does not break. If a quantity of heat equal to the magnitude of the change in mechanical energy of the water goes into the water, what is its increase in temperature?

Step-by-Step Solution

Answer

The temperature increase of the water is approximately 0.527°C.

Step 1 - Understand the Problem

We need to determine the increase in temperature of water when a water bottle falls from a height of 225 m. The energy lost by the bottle is converted to heat, which increases the water's temperature.

Step 2 - Calculate the Potential Energy Loss

The potential energy (PE) lost by the bottle is given by the formula \( PE = mgh \), where \( m \) is mass in kg, \( g \) is the acceleration due to gravity \( 9.8 \, \text{m/s}^2 \), and \( h \) is the height. The volume of the water bottle is 1.00 L, which is equivalent to 1.00 kg. Thus, \( PE = 1.00 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 225 \, \text{m} = 2205 \, \text{J} \).

Step 3 - Calculate Heat Transfer

All the potential energy lost by the falling bottle turns into heat energy. Therefore, the heat energy \( Q \) is equal to the potential energy \( 2205 \, \text{J} \).

Step 4 - Use Heat Energy to Find Temperature Change

The relation between heat \( Q \), mass \( m \), specific heat capacity \( c \), and change in temperature \( \Delta T \) is given by \( Q = mc\Delta T \). For water, \( c = 4186 \, \text{J/kg°C} \). Substitute \( Q = 2205 \, \text{J} \), \( m = 1.00 \, \text{kg} \), and solve for \( \Delta T \): \[ 2205 = 1.00 \times 4186 \times \Delta T \]Solving gives \( \Delta T \approx 0.527 \, \text{°C} \).

Key Concepts

Potential Energy

Potential energy is the energy that is stored in an object due to its position or state. In the context of the exercise, the water bottle held at a height of 225 meters above the ground stores potential energy because of its elevated position. This energy is calculated using the formula:

• \( PE = mgh \)

Here, \( m \) represents the mass of the object, \( g \) is the acceleration due to gravity (approximately 9.8 \( \text{m/s}^2 \) on Earth), and \( h \) is the height above the ground.

When the bottle falls, all this potential energy is expected to convert to other energy forms, primarily mechanical or thermal. Understanding how potential energy works is crucial for analyzing the energy transformation in systems, like the falling water bottle, because it sets the stage for understanding subsequent energy exchanges.

Heat Transfer

When energy changes form, one common conversion is from mechanical energy to thermal energy, a process known as heat transfer. In this exercise, as the potential energy of the water bottle transforms upon hitting the ground, it generates heat. This heat does not leave the system but remains in the water.

• The heat transfer in this scenario occurs because energy is being redirected from the motion and impact of the object into the state of the water at the molecular level.

The primary principle at play here is the conservation of energy, which dictates that energy cannot be created or destroyed, only transformed. Thus, the energy lost in height (potential energy) becomes energy acquired in substance (heat).

Understanding heat transfer is essential in science and engineering, where it can have implications for everything from designing engines and refrigerators to predicting weather patterns.

Specific Heat Capacity

Specific heat capacity is a material-specific property that defines how much heat is needed to change the temperature of 1 kilogram of the substance by 1 degree Celsius. For water, this value is remarkably high at 4186 \( \text{J/kg°C} \), meaning water requires a significant amount of energy to change its temperature.

• This characteristic of water makes it effective in buffering temperature changes, which is why it is often used in heat regulation and cooling systems.
• In the provided exercise, knowing the specific heat capacity allows us to calculate the temperature increase from the heat energy gained due to mechanical energy conversion.

Given that the total heat energy transferred to the water is \( 2205 \, \text{J} \), we can use the equation:

• \( Q = mc\Delta T \)

to solve for the temperature change \( \Delta T \). This equation illustrates how when heat energy \( Q \) is absorbed by an object, its mass \( m \) and specific heat capacity \( c \) will determine the resultant temperature increase \( \Delta T \). Understanding specific heat capacity helps us predict how substances will react to energy inputs, which is vital across numerous scientific and industrial applications.