Problem 2
Question
A swimmer does \(4.3 \times 10^{5} \mathrm{~J}\) of work and gives off \(1.7 \times 10^{5} \mathrm{~J}\) of heat during a workout. Determine \(\Delta E, W\), and \(Q\) for the swimmer.
Step-by-Step Solution
Verified Answer
\(\Delta E = 2.6 \times 10^{5} \text{ J}\), \(W = 4.3 \times 10^{5} \text{ J}\), \(Q = -1.7 \times 10^{5} \text{ J}\)
1Step 1: Identify Given Values
We are given that the swimmer does \(4.3 \times 10^{5} \, \text{J}\) of work and gives off \(1.7 \times 10^{5} \, \text{J}\) in the form of heat. These correspond to work \(W = 4.3 \times 10^{5} \, \text{J}\) and heat \(Q = -1.7 \times 10^{5} \, \text{J}\) (negative because it is energy released).
2Step 2: State the First Law of Thermodynamics
The first law of thermodynamics is given by the formula \(\Delta E = Q + W\), where \(\Delta E\) is the change in internal energy, \(Q\) is the heat exchanged, and \(W\) is the work done.
3Step 3: Substitute Known Values into the Formula
Substitute the given values into the first law of thermodynamics equation: \(\Delta E = (-1.7 \times 10^{5} \text{ J}) + (4.3 \times 10^{5} \text{ J})\).
4Step 4: Perform the Calculation
Calculate \(\Delta E\) by solving \((-1.7 \times 10^{5}) + (4.3 \times 10^{5})\). This results in \((4.3 - 1.7) \times 10^{5}\), which simplifies to \(2.6 \times 10^{5} \, \text{J}\).
5Step 5: Summarize the Solution
The change in internal energy \(\Delta E\) is \(2.6 \times 10^{5} \, \text{J}\), the work done \(W\) is \(4.3 \times 10^{5} \, \text{J}\), and the heat exchanged \(Q\) is \(-1.7 \times 10^{5} \, \text{J}\).
Key Concepts
Work-Energy PrincipleHeat TransferInternal Energy Change
Work-Energy Principle
The work-energy principle is a fundamental concept in physics which states that work done by all forces acting on a system results in a change in that system's energy. Simply put, when energy is applied to an object through work, the object's energy changes.
This principle plays a crucial role in understanding how the swimmer in our exercise exerts a force to move through water. The swimmer does a total of \(4.3 \times 10^{5} \, \text{J}\) of work, which means energy has been transferred from the swimmer's muscles to the water. This energy aids the movement, propelling the swimmer forward.
This principle plays a crucial role in understanding how the swimmer in our exercise exerts a force to move through water. The swimmer does a total of \(4.3 \times 10^{5} \, \text{J}\) of work, which means energy has been transferred from the swimmer's muscles to the water. This energy aids the movement, propelling the swimmer forward.
- Work done by the swimmer equates to energy applied to their body and the water.
- Understanding work helps us calculate energy changes in various situations.
Heat Transfer
Heat transfer refers to the movement of thermal energy from one object or system to another due to a difference in temperature. In relation to the first law of thermodynamics, heat transfer affects the internal energy of a system.
In the swimmer's case, they release \(1.7 \times 10^{5} \, \text{J}\) of heat as their body adjusts to maintain temperature. This release acts as a form of energy exchange, helping keep the swimmer's body cool during exertion.
In the swimmer's case, they release \(1.7 \times 10^{5} \, \text{J}\) of heat as their body adjusts to maintain temperature. This release acts as a form of energy exchange, helping keep the swimmer's body cool during exertion.
- A negative sign indicates that the swimmer is giving off heat.
- Heat transfer can be conductive, convective, or radiative.
Internal Energy Change
Internal energy change, denoted as \(\Delta E\), is a measure of how the total energy within a system varies, mainly caused by heat exchange and work done. According to the first law of thermodynamics, this change in internal energy is expressed through the equation \(\Delta E = Q + W\).
For the swimmer, \(\Delta E\) is calculated by combining the work done and the heat given off:
\[\Delta E = (-1.7 \times 10^{5} \, \text{J}) + (4.3 \times 10^{5} \, \text{J}) = 2.6 \times 10^{5} \, \text{J}\]
For the swimmer, \(\Delta E\) is calculated by combining the work done and the heat given off:
\[\Delta E = (-1.7 \times 10^{5} \, \text{J}) + (4.3 \times 10^{5} \, \text{J}) = 2.6 \times 10^{5} \, \text{J}\]
- The positive \(\Delta E\) implies an increase in the swimmer's total internal energy.
- This energy change is the result of both mechanical work and thermal processes.
Other exercises in this chapter
Problem 1
Follow-up After walking for a few minutes, you begin to run, doing \(5.1 \times 10^{5} \mathrm{~J}\) of work and decreasing your thermal energy by \(8.8 \times
View solution Problem 3
Give the change in thermal energy of a system if (a) \(W=50 \mathrm{~J}, Q=50 \mathrm{~J}\); (b) \(W=-50 \mathrm{~J}, Q=-50 \mathrm{~J}\); or (c) \(W=50 \mathrm
View solution Problem 4
Follow-up What is the efficiency of a heat engine that does \(1250 \mathrm{~J}\) of work and gives off \(5250 \mathrm{~J}\) of heat to the cold reservoir?
View solution Problem 5
What is the efficiency of an engine that exhausts \(870 \mathrm{~J}\) of heat in the process of doing \(340 \mathrm{~J}\) of work?
View solution