Thermodynamics Efficiency
Efficiency in thermodynamics represents how effectively a machine converts input energy into useful work. No engine is
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efficient because some energy is always lost

Question

Thermodynamics Efficiency
Efficiency in thermodynamics represents how effectively a machine converts input energy into useful work. No engine is 
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0
%
 ﻿efficient because some energy is always lost as heat.
Question in Mechanical Engineering
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)
 ﻿A heat engine receives 
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 ﻿J of heat energy and produces 
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4
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 ﻿J of useful work. Calculate the efficiency of the engine and heat rejected to the surroundings.

VIPSolutions · Accepted Answer

# Thermodynamic Efficiency: Example Problem

## Problem Statement

A heat engine receives $ 500 $ J of energy ($ Q_{in} $) and produces  J of useful work ($ W $).  
Calculate:
- The efficiency ($ \eta $) of the engine.
- The heat rejected ($ Q_{out} $) to the surroundings.

---

## 1. Efficiency Calculation

**Efficiency ($ \eta $)** is defined as:
$$
\eta = \frac{\text{Useful Work Output}}{\text{Heat Input}} \times 100\%
$$

Plug in the given values:
$$
\eta = \frac{140\ \text{J}}{500\ \text{J}} \times 100\% = 28\%
$$

---

## 2. Heat Rejected Calculation

The **first law of thermodynamics** gives:
$$
Q_{in} = W + Q_{out}
$$
$$
Q_{out} = Q_{in} - W
$$
$$
Q_{out} = 500\ \text{J} - 140\ \text{J} = 360\ \text{J}
$$

---

## 3. Summary Table

| Quantity         | Value        |
|------------------|-------------|
| Heat Input ($ Q_{in} $)     | 500 J       |
| Useful Work ($ W $)         | 140 J       |
| Efficiency ($ \eta $)       | 28%          |
| Heat Rejected ($ Q_{out} $) | 360 J       |

---

## 4. Key Points

- No real engine is 100% efficient because some energy is always lost as heat.
- The efficiency is always less than 100%.

---

**Image Example:**  
![Heat engine schematic showing Q_in, W, and Q_out](https://upload.wikimedia.org/wikipedia/commons//07/Heat_engine_cycle.png)
*Heat engine schematic: $ Q_{in} $ enters, $ W $ exits as work, $ Q_{out} $ is rejected heat*

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