Problem 3 - University Physics with Modern Physics

Question

A gasoline engine takes in 1.61 \(\times\) 10\(^4\) J of heat and delivers 3700 J of work per cycle. The heat is obtained by burning gasoline with a heat of combustion of 4.60 \(\times\) 10\(^4\) J/g. (a) What is the thermal efficiency? (b) How much heat is discarded in each cycle? (c) What mass of fuel is burned in each cycle? (d) If the engine goes through 60.0 cycles per second, what is its power output in kilowatts? In horsepower?

Step-by-Step Solution

Answer

(a) 22.9% (b) 1.24 × 10⁴ J (c) 0.350 g (d) 222 kW, 297.5 HP

Step 1: Calculate the Thermal Efficiency

The thermal efficiency of an engine is given by the formula \( \eta = \frac{W_{out}}{Q_{in}} \), where \( W_{out} = 3700 \text{ J} \) is the work done and \( Q_{in} = 1.61 \times 10^4 \text{ J} \) is the heat input. Substitute these values to find the thermal efficiency: \( \eta = \frac{3700}{1.61 \times 10^4} = 0.229 \) or 22.9%.

Step 2: Calculate the Heat Discarded

The heat discarded or rejected \( Q_{out} \) can be calculated using the formula \( Q_{out} = Q_{in} - W_{out} \). Substitute the known values: \( Q_{out} = 1.61 \times 10^4 \text{ J} - 3700 \text{ J} = 1.24 \times 10^4 \text{ J} \).

Step 3: Calculate the Mass of Fuel Burned

To find the mass of fuel burned, use the heat of combustion formula \( m = \frac{Q_{in}}{H_c} \), where \( Q_{in} = 1.61 \times 10^4 \text{ J} \) and \( H_c = 4.60 \times 10^4 \text{ J/g} \). Thus, \( m = \frac{1.61 \times 10^4}{4.60 \times 10^4} = 0.350 \text{ g} \).

Step 4: Calculate Power Output in Kilowatts

Power is calculated using \( P = \frac{W_{out} \times ext{cycles per second}}{1000} \). With \( W_{out} = 3700 \text{ J} \) and 60 cycles per second, \( P = \frac{3700 \times 60}{1000} = 222 \text{ kW} \).

Step 5: Convert Power Output to Horsepower

Convert the power output from kilowatts to horsepower using the conversion factor 1 kW = 1.341 HP: \( P_{HP} = 222 \times 1.341 = 297.5 \text{ HP} \).

Key Concepts

Thermal Efficiency

Thermal efficiency is a measure of how well a heat engine converts the heat it receives into useful work. In simple terms, it indicates the engine's effectiveness. For instance, if an engine has a thermal efficiency of 22.9%, it means that 22.9% of the heat energy supplied is converted into work while the rest is often considered wasted or lost. This is a crucial parameter as it helps in evaluating the performance of an engine. To calculate thermal efficiency, we use the formula: \[ \eta = \frac{W_{out}}{Q_{in}} \]where \( W_{out} \) is the output work and \( Q_{in} \) is the input heat energy. For example, if \( W_{out} = 3700 \text{ J} \) and \( Q_{in} = 1.61 \times 10^4 \text{ J} \), we find the thermal efficiency by substituting these values into the formula. This results in \( \eta = 0.229 \) or 22.9%. Understanding thermal efficiency can help in designing more efficient engines by reducing the amount of energy lost.

Heat Engine

A heat engine is a system that converts heat or thermal energy into mechanical energy, which can then be used to perform work. Heat engines often operate in a cycle, repeatedly taking in heat, converting it into work, and then expelling the excess heat. There are many types of heat engines, including gasoline engines, steam engines, and even the human body to some extent. The functioning of a heat engine can be understood by its cycle:

• In the first phase, the engine absorbs a certain amount of heat \( (Q_{in}) \).
• Then, it converts part of this absorbed heat into work \( (W_{out}) \).
• The remaining heat, which is not converted, is discarded \( (Q_{out}) \).

The efficiency and the work output of a heat engine are crucial in determining how effective it is in its operation. Understanding the concepts of heat conversion and energy cycles in such engines can lead to more innovative and highly efficient engine designs.

Power Output

Power output is an important concept in thermodynamics. It refers to the amount of work an engine performs over a certain period. Usually, it tells us how powerful an engine is because it combines the rate of energy conversion with the total amount of energy converted. This is essential for understanding the capacity of engines, machines, and systems. The formula for calculating power output is:\[ P = \frac{W_{out} \times \text{cycles per second}}{1000} \]For example, using \( W_{out} = 3700 \text{ J} \) and 60 cycles per second, one can calculate the power output in kilowatts as \( P = 222 \text{ kW} \). Additionally, to convert this into horsepower, a common unit, we can use:\[ P_{HP} = P \times 1.341 \]This conversion helps people understand and compare engines and motors easily. Understanding power output helps in analyzing the effectiveness and capacity of different engines and mechanical systems.

Mass of Fuel Burned

The mass of fuel burned in a cycle of an engine is a key factor in understanding its efficiency and environmental impact. It tells us how much fuel is consumed, which affects not only cost but also emissions. Knowing the exact fuel mass helps in optimizing engine performance and developing more sustainable energy solutions. The formula used to find the mass of fuel burned, given the heat of combustion, is:\[ m = \frac{Q_{in}}{H_{c}} \]Here, \( Q_{in} \) is the total heat energy input and \( H_{c} \) is the heat of combustion of the fuel. For instance, with \( Q_{in} = 1.61 \times 10^4 \text{ J} \) and \( H_{c} = 4.60 \times 10^4 \text{ J/g} \), we find \( m = 0.350 \text{ g} \) of fuel burned per cycle.Understanding and reducing the mass of fuel burned is crucial for creating more efficient and environmentally friendly engines.