/*! 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 32 While painting the top of an ant... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

While painting the top of an antenna 225 \(\mathrm{m}\) in height, a worker accidentally lets a \(1.00-\mathrm{L}\) wattle 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?

Short Answer

Expert verified
The water's temperature increases by approximately 0.527°C.

Step by step solution

01

Understand the Problem

The problem involves a change in mechanical energy, specifically gravitational potential energy, as the water bottle falls from 225 meters to the ground. This change in mechanical energy is converted into heat, which then increases the temperature of the water in the bottle.
02

Calculate the Change in Mechanical Energy

Mechanical energy change is equal to the change in gravitational potential energy, which is calculated using the formula: \[ \Delta E = mgh \] where \( m \) is the mass of the water, \( g \) is the acceleration due to gravity (\(9.81\, \text{m/s}^2\)), and \( h \) is the height (225 m). Since the density of water is \(1\, \text{kg/L}\), the mass \( m \) of 1 L of water is 1 kg. Thus, \[ \Delta E = 1 \times 9.81 \times 225 = 2207.25 \text{ J} \].
03

Relate to Temperature Change

The heat energy transferred to the water \( Q \) is equal to the change in mechanical energy calculated. The increase in temperature \( \Delta T \) can be found using the specific heat capacity formula: \[ Q = mc\Delta T \] where \( c \) is the specific heat capacity of water \(4.186 \text{ J/g°C}\), and \( m \) is 1000 g (since 1 kg = 1000 g). Solving for \( \Delta T \):\[ \Delta T = \frac{Q}{mc} = \frac{2207.25}{1000 \times 4.186} = 0.527 \text{°C} \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Gravitational Potential Energy
Gravitational Potential Energy (GPE) is a form of energy an object possesses because of its position in a gravitational field. Essentially, it's the energy stored in an object as it is positioned above a reference level, such as the ground. The formula to calculate the gravitational potential energy is \[ GPE = mgh \]. Here, \( m \) represents mass, \( g \) is the acceleration due to gravity (approximately \(9.81\, \text{m/s}^2\) on Earth), and \( h \) stands for the height above the reference level. The GPE highlights two essential components:
  • **Mass of the Object**: A heavier object contains more potential gravitational energy when at the same height as a lighter one.
  • **Height of the Object**: The higher the position of the object above the ground, the more gravitational potential energy it possesses.
When a falling object reaches the ground, its gravitational potential energy is converted largely to other forms of energy, such as heat. This transformation of energy is seen in the exercise, where the energy change contributes to an increase in the water bottle's temperature.
Temperature Change
Temperature change in an object happens when energy is absorbed or released, influencing the object's thermal state. The rise or fall in temperature depends largely on the heat energy transferred, the object's mass, and its specific heat capacity. In the context of the exercise, the temperature change occurs as the gravitational potential energy is converted into heat when the water bottle hits the ground. The amount of temperature shift \( \Delta T \) is determined using the formula:\[ \Delta T = \frac{Q}{mc} \]Here, \( Q \) is the heat energy, \( m \) is the mass, and \( c \) symbolizes the specific heat capacity. A clear understanding of this equation helps in discerning how external factors, like energy transfer, impact temperature variations. In many scenarios:
  • **Higher the energy transfer**, the more significant the temperature change will be.
  • **Mass of the object**: More massive objects experience less temperature change for the same amount of energy transferred.
This principle is crucial in understanding the energy transformations in both everyday occurrences and more complex scientific processes.
Specific Heat Capacity
Specific heat capacity is a property that describes the amount of heat required to change a substance's temperature by a given amount, typically 1°C. It's a pivotal factor in determining how substances react to thermal energy. In the equation \( Q = mc\Delta T \), \( c \) stands for the specific heat capacity, which plays a significant role in calculating temperature change. The specific heat capacity of water is notably high at \(4.186 \text{ J/g°C}\), meaning it takes more energy to alter water's temperature compared to many other substances. Key insights about specific heat capacity include:
  • **Substances with higher specific heat** efficiently resist temperature changes when energy is added or removed.
  • **Water's high specific heat capacity** gives it excellent thermal stability, making it essential in regulating climates and supporting life.
Understanding specific heat helps in numerous practical scenarios, like cooking, where knowing how a material will respond to heating can guide temperature control for optimal results.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) A wire that is 1.50 \(\mathrm{m}\) long at \(20.0^{\circ} \mathrm{C}\) is found to increase in length by 1.90 \(\mathrm{cm}\) when warmed to \(420.0^{\circ} \mathrm{C}\) . Compute its average coefficient of linear expansion for this temperature range. (b) The wire is stretched just taut (zero tension) at \(420.0^{\circ} \mathrm{C}\). Find the stress in the wire if it is cooled to \(20.0^{\circ} \mathrm{C}\) without being allowed to contract. Young's modulus for the wire is \(2.0 \times 10^{11} \mathrm{Pa}\)

While running, a 70 -kg student generates thermal energy at a rate of 1200 \(\mathrm{W}\) . For the runner to maintain a constant body temperature of \(37^{\circ} \mathrm{C},\) this energy must be removed by perspiration or other mechanisms. If these mechanisms failed and the heat could not flow out of the student's body, for what amount of time could a student run before irreversible body damage occurred? (Note: Protein structures in the body are irreversibly damaged if body temperature rises to \(44^{\circ} \mathrm{C}\) or higher. The specific heat of a typical human body is \(3480 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K},\) slightly less than that of water. The difference is due to the presence of protein, fat,and minerals, which have lower specific heats.)

Food Intake of a Hamster. The energy output of an animal engaged in an activity is called the basal metabolic rate \((\mathrm{BMR})\) and is a measure of the conversion of food energy into other forms of energy. A simple calorimeter to measure the BMR consists of an insulated box with a thermometer to measure the temperature of the air. The air has density 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) and specific heat \(1020 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K} . \mathrm{A} 50.0-\mathrm{g}\) hamster is placed in a calorimeter that contains 0.0500 \(\mathrm{m}^{3}\) of air at room temperature. (a) When the hamster is running in a wheel, the temperature of the air in the calorimeter rises 1.60 \(\mathrm{C}^{\circ}\) per hour. How much heat does the running hamster generate in an hour? Assume that all this heat goes into the air in the calorimeter. You can ignore the heat that goes into the walls of the box and into the thermometer, and assume that no heat is lost to the surroundings. (b) Assuming that the hamster converts seed into heat with an efficiency of 10\(\%\) and that hamster seed has a food energy value of \(24 \mathrm{J} / \mathrm{g},\) how many grams of seed must the hamster eat per hour to supply this energy?

(a) If an area measured on the surface of a solid body is \(A_{0}\) at some initial temperature and then changes by \(\Delta A\) when the temperature changes by \(\Delta T,\) show that $$\Delta A=(2 \alpha) A_{0} \Delta T$$ where \(\alpha\) is the coefficient of linear expansion. (b) A circular sheet of aluminum is 55.0 \(\mathrm{cm}\) in diameter at \(15.0^{\circ} \mathrm{C}\) . By how much does the area of one side of the sheet change when the temperature increases to \(27.5^{\circ} \mathrm{C}\) ?

A copper calorimeter can with mass 0.446 \(\mathrm{kg}\) contains 0.0950 \(\mathrm{kg}\) of ice. The system is initially at \(0.0^{\circ} \mathrm{C} .\) (a) If 0.0350 \(\mathrm{kg}\) of steam at \(100.0^{\circ} \mathrm{C}\) and 1.00 atm pressure is added to the can, what is the final temperature of the calorimeter can and its contents? (b) At the final temperature, how many kilograms are there of ice, how many of liquid water, and how many of steam?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.