Problem 75 - Pearson Physics

Question

An engine receives \(780 \mathrm{~J}\) of heat from a hot reservoir and does \(110 \mathrm{~J}\) of work. What is (a) the heat given off to the cold reservoir and (b) the efficiency of this engine?

Step-by-Step Solution

Answer

The heat given off is 670 J and the engine's efficiency is 14.1%.

Step 1: Understanding the Problem

We need to find two things for this engine: (a) the heat given off to the cold reservoir and (b) the efficiency. We know that the engine receives 780 J of heat from the hot reservoir and does 110 J of work.

Step 1: Applying the First Law of Thermodynamics

The first law of thermodynamics states that the total energy input into the system equals the work done plus the energy expelled. Mathematically, this is \( Q_h = W + Q_c \), where \( Q_h \) is the heat input, \( W \) is the work done, and \( Q_c \) is the heat expelled to the cold reservoir. We know \( Q_h = 780 \) J and \( W = 110 \) J.

Step 2: Calculate the Heat Given Off to the Cold Reservoir

Using the equation from Step 1, substitute the known values: \( 780 = 110 + Q_c \). Rearrange to find \( Q_c \): \( Q_c = 780 - 110 = 670 \) J. Thus, the engine gives off 670 J of heat to the cold reservoir.

Step 3: Calculate the Efficiency of the Engine

The efficiency \( \, \eta \, \) of a heat engine is defined as the ratio of the work done by the engine to the heat input: \( \eta = \frac{W}{Q_h} \times 100\% \). Substitute the known values: \( \eta = \frac{110}{780} \times 100\% = 14.1\% \). Thus, the efficiency of the engine is 14.1\%.

Key Concepts

Engine Efficiency

Engine efficiency measures how well an engine converts the heat from a hot reservoir into useful work. It's a crucial concept, especially when dealing with heat engines like car engines, power plants, or even refrigerators.
Understanding efficiency involves calculating the ratio of the work done by the engine to the heat energy supplied to it. Mathematically, this is expressed as:

• \[\eta = \frac{W}{Q_h} \times 100\%\]

where \(W\) is the work output and \(Q_h\) is the heat input.
In our exercise, with 780 J of heat input and 110 J of work output, the engine's efficiency is 14.1%.
Efficiency is always less than 100% due to inevitable energy losses, such as heat dissipation, making it impossible for a real-world engine to convert all heat into work.

Heat Transfer

Heat transfer is a fundamental process in thermodynamics, involving the movement of thermal energy from one place to another. In a heat engine, such as the one in our exercise, heat transfers from a hot reservoir to the engine.
This energy is partially converted into work, and the rest is expelled as waste heat to a cold reservoir.
In our problem, the engine absorbs 780 J of heat from the hot reservoir.
This process can be analytically described by the first law of thermodynamics, which dictates:

• \[Q_h = W + Q_c\]

Where \(Q_h\) is the heat absorbed, \(W\) is the work done (110 J), and \(Q_c\) is the heat given off to the cold reservoir.
By rearranging this formula, we find that 670 J of heat is transferred to the cold reservoir.

Thermodynamics Problem Solving

Solving thermodynamics problems relies heavily on understanding and applying core principles like the first law of thermodynamics.
To tackle problems efficiently, follow a structured approach:
1. **Understand** the problem and identify known values.
2. **Select** relevant thermodynamic principles or formulas that apply.
3. **Substitute** the known values and solve for the unknowns.
In our exercise, we used the first law to set up the equation:

• \[780 = 110 + Q_c\]

Solving for \(Q_c\), we discovered the engine releases 670 J as waste heat.
Then, to find the efficiency, we calculated \(\eta\) using given energy values.
Always double-check calculations and understand the underlying physics to ensure accuracy.