Question

The useful energy that comes out of an energy transfer process is related to the efficiency of the process by the following equation: $$ \begin{aligned} &\text { total } \\ &\text { consumed } \end{aligned} \times \text { efficiency }=\begin{aligned} &\text { useful } \\ &\text { energy } \end{aligned} $$ where the efficiency is in decimal (not percent) form. a. If a process is \(30 \%\) efficient, how much useful energy can be derived if \(455 \mathrm{~kJ}\) are consumed? b. A person eats approximately \(2200 \mathrm{kcal} /\) day. How much of that energy is available to do physical work? c. If a car needs \(5.0 \times 10^{3} \mathrm{~kJ}\) to go a particular distance, how much energy will be consumed if the car is \(20 \%\) efficient? d. If an electrical power plant produces \(1.0 \times 10^{9} \mathrm{~J}\) of electrical energy, how much energy will be consumed by the plant if it is \(34 \%\) efficient?

Short answer

a. 136.5 kJ; b. Assuming 22.5% efficiency, the useful energy is about 2200 kcal \times 4.184 kJ/kcal \times 0.225 = 2.18 x 10^4 kJ; c. 2.5 x 10^4 kJ; d. 2.94 x 10^9 J

Step-by-step solution

Convert Percentage Efficiency to Decimal

For part a, the efficiency is given as 30%. To use it in calculations, convert it to decimal by dividing by 100. This gives an efficiency of 0.30.

Calculate Useful Energy for Part a

Use the given equation \[\text{total consumed} \times \text{efficiency} = \text{useful energy}\]with 455 kJ of energy consumed and an efficiency of 0.30 to find the useful energy. Multiply 455 kJ by 0.30 to get the useful energy.

Determine Efficiency for Daily Energy Intake

For part b, assume a typical efficiency rate at which the human body converts food into energy that can be used for physical work. This rate is typically about 20-25%. For this solution, use an average value of 22.5% unless otherwise given. Convert this to decimal form which gives us 0.225.

Convert Daily Energy Intake from kcal to kJ

Before calculating the useful energy from the daily intake, convert the energy from kcal to kJ. Use the conversion factor that 1 kcal = 4.184 kJ. Multiply 2200 kcal by 4.184 kJ/kcal.

Calculate Useful Energy for Part b

Now calculate the useful energy from the daily intake using the equation, with the converted energy intake in kJ and the efficiency in decimal form.

Calculate Energy Consumed for Part c

Now reverse the equation for part c, where we need to find total energy consumed given the useful energy and the efficiency. The car needs 5.0 x 10^3 kJ of useful energy and operates at 20% efficiency. Convert the efficiency to decimal form, 0.20, and divide the useful energy by this decimal to find the total consumption.

Calculate Energy Consumed for Part d

For part d, use the equation to find the total energy consumed by the power plant. It produces 1.0 x 10^9 J of electrical energy at 34% efficiency. Convert the efficiency to decimal, 0.34, and divide the electrical energy produced by this decimal to find the total energy consumed by the plant.

Key concepts

Thermodynamics

Thermodynamics is a branch of physics that deals with heat, work, and temperature, and their relation to energy, radiation, and physical properties of matter. One of the key principles of thermodynamics is the conservation of energy, which states that energy cannot be created or destroyed, only transferred or transformed from one form to another.

This core principle is directly related to the exercises at hand, where the students are asked to determine the useful energy output of various processes. Understanding the efficiency of energy conversion in these processes requires an understanding of thermodynamic principles. For example, heat engines, which may be in a car or a thermal power plant, convert thermal energy into work with a certain efficiency determined by thermodynamic limits such as the Carnot efficiency.

When discussing energy efficiency in the context of thermodynamics, it is important to consider not just the input and output energy forms but also the waste heat and the role of entropy, which is a measure of disorder or randomness in a system. Higher entropy typically means less ability to do useful work, and thus, energy becomes less useful.

Energy Transfer

Energy transfer refers to the movement of energy from one place or physical system to another. There are different methods of energy transfer, including conduction, convection, radiation, and through mechanical work. Knowing how to calculate energy transfer is central to solving problems dealing with efficiency.

In the given exercise, students are required to understand that the consumed energy must first be converted into other forms, like mechanical or electrical energy, before it can be used. The efficiency of this process is crucial because it dictates how much of the consumed energy can actually be used to perform work or produce a useful outcome.

For instance, in the case of the car mentioned in the exercise, the chemical energy contained in the fuel (consumed energy) is transformed into kinetic energy to move the car. However, only a fraction of the initial energy serves this purpose; the rest is lost—often as heat. That fraction that does useful work is what is targeted in calculating energy transfer.

Units of Energy Conversion

In energy conversion calculations, it is essential to be cognizant of the different units that may be used to measure energy. Common units include joules (J), kilojoules (kJ), and kilocalories (kcal), among others. One must be proficient in converting these units to maintain consistency and accuracy in calculations.

In the steps of the provided solution, there's a need to convert kcal to kJ to use with the efficiency formula. The conversion factor is a critical piece of information: 1 kcal equals 4.184 kJ. Without proper conversion, the analysis of energy efficiency could yield incorrect results.

Furthermore, it's important to remember that the specific energy content of various substances might be measured in different units. For instance, food energy is commonly given in calories or kilocalories, whereas mechanical or electrical energy is typically measured in joules or kilojoules.

Scientific Problem-Solving

Scientific problem-solving is a systematic approach used to understand complex scenarios and determine solutions using scientific methods. This involves defining the problem, gathering data, formulating a hypothesis, testing that hypothesis, and analyzing results.

Applying this method to the given exercise, we first define the objective (calculating useful energy output or total energy consumption). Then we gather necessary data which includes efficiency rates and consumed energy values. The hypothesis in such problems is typically a formula or principle that predicts the output – like the efficiency equation given in the exercise.

The testing phase involves performing mathematical operations correctly, ensuring all variables are accounted for, like converting units where necessary. Lastly, analyzing results is about interpreting the calculated values and understanding their practical implications – such as how efficient a car is or how much of the food energy we consume is actually available for physical activity.

• It’s crucial to ensure that assumptions are stated clearly, as they can influence the direction and validity of the problem-solving process. For example, the assumption of the human body's efficiency rate.
• Documentation of each step provides clarity and promotes a better understanding, allowing for more accurate problem-solving.