Problem 50 - Pearson Physics

Question

An engine receives \(770 \mathrm{~J}\) of heat from a hot reservoir and does \(160 \mathrm{~J}\) of work. What is (a) the efficiency of this engine and (b) the heat given off to the cold reservoir?

Step-by-Step Solution

Answer

(a) The efficiency is approximately 20.78%. (b) The heat given off to the cold reservoir is 610 J.

Step 1: Understanding the Problem

When an engine receives heat energy from a hot reservoir, it uses part of this energy to do work and dissipates the rest to a cold reservoir. We need to find the efficiency of the engine and the heat that is given off to the cold reservoir. We know the heat supplied is \(Q_H = 770 \, \text{J}\) and the work done \(W = 160 \, \text{J}\).

Step 1: Calculate Efficiency

Efficiency (\(\eta\)) of an engine is given by the formula \(\eta = \frac{W}{Q_H}\), where \(W\) is the work done and \(Q_H\) is the heat input. Substituting the given values, we have: \[\eta = \frac{160 \, \text{J}}{770 \, \text{J}}\approx 0.2078.\] To express it as a percentage, we multiply by 100: \[\eta \approx 20.78\%.\]

Step 2: Calculate Heat Given Off to Cold Reservoir

The heat given off to the cold reservoir (\(Q_C\)) can be calculated using the formula \(Q_C = Q_H - W\). Substituting the known values, we have: \[Q_C = 770 \, \text{J} - 160 \, \text{J} = 610 \, \text{J}.\]

Key Concepts

Heat Engine

A heat engine is a system that converts heat or thermal energy to mechanical work. This is a central concept in thermodynamics, often found in engines that power cars and other machines. Heat engines work by absorbing energy from a high-temperature source, using a part of this energy to perform work, and releasing any leftover energy to a cold reservoir.
A typical heat engine might include:

• A hot reservoir that provides energy.
• A working substance that converts energy to work, like a piston or turbine.
• A cold reservoir that absorbs excess energy.

The efficiency of these engines is never 100% because not all absorbed energy can be converted into useful work due to the second law of thermodynamics.

Efficiency Calculation

The efficiency of a heat engine (\(\eta\)) is a measure of how well it converts heat into work. It's crucial for determining the performance of engines and the viability of engineering solutions.
Efficiency is calculated using the ratio of work done (\(W\)) to the heat absorbed from the hot reservoir (\(Q_H\)).

The formula for efficiency is:\[\eta = \frac{W}{Q_H}\]
This ratio gives the efficiency as a decimal, which can be converted into a percentage by multiplying by 100. For example, if a heat engine does 160 Joules of work from 770 Joules of heat input, the efficiency is about 20.78% using:\[\eta = \frac{160}{770} \approx 0.2078 (20.78\%)\]Efficient engines maximize the amount of work done per unit of heat absorbed, making them highly valuable in conserving energy and reducing waste.

Heat Transfer

Heat transfer is the movement of thermal energy from one object or substance to another. In the context of a heat engine, it's the process through which energy moves from the hot reservoir to the working substance and then to the cold reservoir.
Understanding heat transfer involves three primary mechanisms:

• Conduction: Transfer of heat through a material without any movement of the material itself.
• Convection: Transfer of heat by the physical movement of a fluid (liquid or gas).
• Radiation: Transfer of heat in the form of electromagnetic waves, such as sunlight.

In a heat engine, heat transfer is critical to the engine's efficiency. After the engine does work, the remaining energy is often transferred to a cold reservoir (or surroundings), which is essential to completing the cycle of a heat engine. The heat rejected to the cold reservoir is calculated by subtracting the work done from the heat absorbed, like in the example: \[Q_C = Q_H - W = 770 ext{ J} - 160 ext{ J} = 610 ext{ J}\]This rejected heat must be managed effectively to maintain the engine's performance and prevent overheating.