/*! 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 101 An object of mass \(6.00 \mathrm... [FREE SOLUTION] | 91Ó°ÊÓ

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An object of mass \(6.00 \mathrm{~kg}\) falls through a height of \(50.0 \mathrm{~m}\) and, by means of a mechanical linkage, rotates a paddle wheel that stirs \(0.600 \mathrm{~kg}\) of water. Assume that the initial gravitational potential energy of the object is fully transferred to thermal energy of the water, which is initially at \(15.0^{\circ} \mathrm{C}\). What is the temperature rise of the water?

Short Answer

Expert verified
The temperature rise of the water is approximately 1.17°C.

Step by step solution

01

Calculate Gravitational Potential Energy (GPE)

The gravitational potential energy lost by the object is given by the formula: \( ext{GPE} = mgh \), where \( m \) is the mass of the object (6.00 kg), \( g \) is the acceleration due to gravity (9.81 m/s²), and \( h \) is the height (50.0 m). So, \( \text{GPE} = 6.00 \times 9.81 \times 50.0 \).
02

Evaluate GPE

Calculate the GPE from Step 1: \( \text{GPE} = 6.00 \times 9.81 \times 50.0 = 2943.0 \) joules.
03

Convert GPE to Thermal Energy

Assume that all the gravitational potential energy is converted into thermal energy for the water. Therefore, the thermal energy added to the water \( Q \) is 2943.0 joules.
04

Calculate Temperature Rise

Use the formula for temperature rise \( Q = mc\Delta T \), where \( Q \) is the thermal energy (2943.0 J), \( m \) is the mass of the water (0.600 kg), and \( c \) is the specific heat capacity of water (approximately 4186 J/kg°C). Rearrange the formula to find \( \Delta T \): \( \Delta T = \frac{Q}{mc} = \frac{2943.0}{0.600 \times 4186} \).
05

Compute \( \Delta T \)

Evaluate the temperature rise: \( \Delta T = \frac{2943.0}{0.600 \times 4186} \approx 1.17 \)°C.

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field. This potential energy can be calculated with the formula: \[ \text{GPE} = mgh \]where:
  • m is the mass of the object in kilograms (kg).
  • g is the acceleration due to gravity, which is approximately 9.81 meters per second squared (m/s²) on Earth.
  • h is the height in meters (m) from which the object falls.
In the given problem, an object with a mass of 6 kg falls from a height of 50 m. By substituting these values into the formula, we calculate the GPE:\[\text{GPE} = 6.00 \times 9.81 \times 50.0 = 2943.0 \text{ joules}.\] The gravitational potential energy initially stored due to the object's height is converted into other types of energy as it descends.
Specific Heat Capacity
Specific heat capacity is a physical property of materials that refers to the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius (°C). It tells us how much energy a substance can absorb before its temperature changes.For water, the specific heat capacity is quite high:
  • Approximately 4186 joules per kilogram per degree Celsius (J/kg°C).
This high value means water can absorb substantial amounts of heat with only a slight change in temperature, which makes it a good medium for thermal energy storage and transfer.In the problem, water with a mass of 0.600 kg is heated. The relationship between the heat energy added to the water and the resulting temperature change is given by:\[ Q = mc\Delta T \]where:
  • \( Q \) is the heat energy added (joules), here 2943.0 J, equal to the GPE of the falling object.
  • \( m \) is the mass of water (0.600 kg).
  • \( c \) is the specific heat capacity (4186 J/kg°C).
  • \( \Delta T \) is the change in temperature (°C).
Rearranging this equation to find \( \Delta T \), we have:\[ \Delta T = \frac{Q}{mc} \approx \frac{2943.0}{0.600 \times 4186} \approx 1.17 \text{°C} \].Thus, the temperature of the water increases by approximately 1.17°C.
Energy Conversion
Energy conversion is the process of changing one form of energy into another. In this exercise, gravitational potential energy is converted into thermal energy as the object falls. When the object falls, the potential energy it holds at height is transformed into kinetic energy, and then into thermal energy when it is used to stir the water.
  • This transformation is facilitated through a mechanical linkage system that connects the falling object to the paddle wheel stirring the water, converting mechanical motion into heat.
  • The full conversion of gravitational potential energy to thermal energy is an example of energy conservation, where the total energy before and after the process remains the same.
Understanding energy conversion principles helps us appreciate how energy can be harnessed and utilized in different systems, highlighting the importance of energy conservation and efficiency in practical applications.

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Most popular questions from this chapter

A vertical glass tube of length \(L=1.280000 \mathrm{~m}\) is half filled with a liquid at \(20.000000^{\circ} \mathrm{C}\). How much will the height of the liquid column change when the tube and liquid are heated to \(30.000000^{\circ} \mathrm{C} ?\) Use coefficients \(\alpha_{\text {glass }}=1.000000 \times 10^{-5} / \mathrm{K}\) and \(\beta_{\text {liquid }}=4.000000 \times 10^{-5} / \mathrm{K}\) .

Suppose that on a linear temperature scale \(X\), water boils at \(-53.5^{\circ} \mathrm{X}\) and freezes at \(-170^{\circ} \mathrm{X}\). What is a temperature of \(340 \mathrm{~K}\) on the \(\mathrm{X}\) scale? (Approximate water's boiling point as \(373 \mathrm{~K}\).)

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