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Chapter 15: Problem 2

A jogger's internal energy changes because he performs \(6.4 \times 10^{5} \mathrm{~J}\) of work and gives off \(4.9 \times 10^{5} \mathrm{~J}\) of heat. However, to cause the same change in his internal energy while walking, he must do \(8.2 \times 10^{5} \mathrm{~J}\) of work. Determine the magnitude of the heat given off while walking.

Short Answer

Expert verified
The heat given off while walking is \( 6.7 \times 10^{5} \) J.

Step by step solution

01

Understanding the Problem

The problem states that a jogger does a certain amount of work and gives off heat, which changes his internal energy. We are given values for work and heat while jogging, and we need to find out how much heat is given off while walking when a different amount of work is done.
02

Using the First Law of Thermodynamics

The first law of thermodynamics states that the change in internal energy (\( \Delta U \)) is the sum of the heat exchanged with the surroundings (\( Q \)) and the work done on or by the system (\( W \)). Mathematically, this is: \( \Delta U = Q - W \).
03

Calculate Change in Internal Energy While Jogging

For jogging, we have the values: \( W = 6.4 \times 10^{5} \) J and \( Q = 4.9 \times 10^{5} \) J. Substituting these into the formula gives: \( \Delta U = 4.9 \times 10^{5} - 6.4 \times 10^{5} \). Calculating, we find \( \Delta U = -1.5 \times 10^{5} \) J.
04

Set Up the Equation for Walking

For walking, we need \( \Delta U \) to be the same as jogging, \(-1.5 \times 10^{5} \) J, and we know that \( W = 8.2 \times 10^{5} \) J. We need to find \( Q \) such that \( \Delta U = Q - W \). So, \(-1.5 \times 10^{5} = Q - 8.2 \times 10^{5} \).
05

Solve for Heat Given Off While Walking

Rearranging the equation from Step 4 gives \( Q = -1.5 \times 10^{5} + 8.2 \times 10^{5} \). Solving this, \( Q = 6.7 \times 10^{5} \) J. This is the amount of heat given off while walking to achieve the same change in internal energy.

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Key Concepts

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

Internal Energy Change
The concept of internal energy change is pivotal in understanding thermodynamics. Internal energy is the total energy contained within a system, encompassing both kinetic and potential energy at the microscopic level.
This includes the random motion of molecules and the forces between them. The change in internal energy, denoted as \( \Delta U \), is the outcome of both heat exchange with the environment and work being performed by or on the system.
In our exercise, the jogger's internal energy change is calculated using the given values.
  • While jogging: \( \Delta U = 4.9 \times 10^{5} - 6.4 \times 10^{5} = -1.5 \times 10^{5} \) J.
  • While walking: the same change \( \Delta U = -1.5 \times 10^{5} \) J is achieved, but the value of work changes, thereby altering the amount of heat exchanged.
Understanding this balance is crucial for analyzing thermodynamic processes.
Work and Heat
Work and heat are the primary ways through which energy is transferred in and out of a system. They play a central role in determining the internal energy of a system related to thermodynamics.
The amount of work done is associated with force being applied over a distance. In our scenario, the jogger performs work measured in joules, such as \( 6.4 \times 10^{5} \) J during jogging.
Heat (\( Q \) ), on the other hand, involves the transfer of energy due to temperature differences. While jogging, the heat given off is \( 4.9 \times 10^{5} \) J.
  • Work done \( W \) and heat \( Q \) affect the internal energy change as described by the equation \( \Delta U = Q - W \).
  • For walking, recalculating heat is essential when given a different work level (\( 8.2 \times 10^{5} \) J).
The interaction between these forms of energy is the essence of the first law of thermodynamics.
Thermodynamic Processes
Thermodynamic processes explore how systems transfer energy and change with interactions involving heat and work. These processes can be isothermal, adiabatic, isochoric, and beyond, each with its defining characteristics.
In any thermodynamic process, like jogging or walking in our problem, energy conservation remains a key principle.
For the jogger's example, both activities represent thermodynamic processes where the energy state changes due to differing levels of work and heat.
  • Jogging: involves a specific balance of work and resulting heat exchange leading to an internal energy change.
  • Walking: also undergoes a thermodynamic change, calculated to maintain the same \( \Delta U \) despite varying work levels.
By understanding these processes, we can predict changes in internal energy and adjust either work or heat to achieve desired outcomes. This understanding aids in applying thermodynamics to real-world scenarios effectively.

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Most popular questions from this chapter

Consider three engines that each use \(1650 \mathrm{~J}\) of heat from a hot reservoir (temperature \(=550 \mathrm{~K}\) ). These three engines reject heat to a cold reservoir (temperature \(=330 \mathrm{~K}\) ). Engine I rejects \(1120 \mathrm{~J}\) of heat. Engine II rejects \(990 \mathrm{~J}\) of heat. Engine III rejects \(660 \mathrm{~J}\) of heat. One of the engines operates reversibly, and two operate irreversibly. However, of the two irreversible engines, one violates the second law of thermodynamics and could not exist. For each of the engines determine the total entropy change of the universe, which is the sum of the entropy changes of the hot and cold reservoirs. On the basis of your calculations, identify which engine operates reversibly, which operates irreversibly and could exist, and which operates irreversibly and could not exist.

Multiple-Concept Example 6 deals with the same concepts as this problem does. What is the efficiency of a heat engine that uses an input heat of \(5.6 \times 10^{4} \mathrm{~J}\) and rejects \(1.8 \times 10^{4} \mathrm{~J}\) of heat?

The inside of a Carnot refrigerator is maintained at a temperature of \(277 \mathrm{~K},\) while the temperature in the kitchen is \(299 \mathrm{~K}\). Using \(2500 \mathrm{~J}\) of work, how much heat can this refrigerator remove from its inside compartment?

A power plant taps steam superheated by geothermal energy to \(505 \mathrm{~K}\) (the temperature of the hot reservoir) and uses the steam to do work in turning the turbine of an electric generator. The steam is then converted back into water in a condenser at 323 \(\mathrm{K}\) (the temperature of the cold reservoir), after which the water is pumped back down into the earth where it is heated again. The output power (work per unit time) of the plant is 84000 kilowatts. Determine (a) the maximum efficiency at which this plant can operate and (b) the minimum amount of rejected heat that must be removed from the condenser every twenty-four hours.

In exercising, a weight lifter loses \(0.150 \mathrm{~kg}\) of water through evaporation, the heat required to evaporate the water coming from the weight lifter's body. The work done in lifting weights is \(1.40 \times 10^{5} \mathrm{~J}\). (a) Assuming that the latent heat of vaporization of perspiration is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg},\) find the change in the internal energy of the weight lifter. (b) Determine the minimum number of nutritional Calories of food ( 1 nutritional Calorie \(=4186 \mathrm{~J}\) ) that must be consumed to replace the loss of internal energy.
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