A motor receives$450\mathrm{J}$of electrical energy from a power supply.$20\%$of this energy is wasted. Find the efficiency of the motor.

Answer: We can first write the equation for heat transfer efficiency as:

$\mathrm{heat}\mathrm{transfer}\mathrm{efficiency}=\frac{\mathrm{useful}\mathrm{heat}\mathrm{output}}{\mathrm{total}\mathrm{heat}\mathrm{input}}\times 100\%\phantom{\rule{0ex}{0ex}}$

We find the amount of useful energy output by subtracting the total energy input of$450\mathrm{J}$by the amount of wasted energy.

$\begin{array}{rcl}\mathrm{useful}\mathrm{heat}\mathrm{output}& =& 450-(450\times \frac{20}{100})\\ & =& 450-90\\ & =& 360\mathrm{J}\end{array}$

Finally, we substitute the useful energy output and total energy output into the efficiency equation.

$\begin{array}{rcl}\mathrm{heat}\mathrm{transfer}\mathrm{efficiency}& =& \frac{\mathrm{useful}\mathrm{heat}\mathrm{output}}{\mathrm{total}\mathrm{heat}\mathrm{input}}\times 100\%\\ & =& \frac{360}{450}\times 100\%\\ & =& 80\%\end{array}$

Q. Find the energy wasted in a system if the useful heat ouput is$350\mathrm{J}$while the total heat input is$900\mathrm{J}$. What is the efficiency of the system?

A. To find the wasted energy we subtract the useful heat output from the total heat input. This gives us$550\mathrm{J}$as shown below.

$\mathrm{Wasted}\mathrm{Energy}=900\mathrm{J}-350\mathrm{J}=550\mathrm{J}$

Next, to find the heat transfer efficiency we use the formula.

$\begin{array}{rcl}\mathrm{heat}\mathrm{transfer}\mathrm{efficiency}& =& \frac{\mathrm{useful}\mathrm{heat}\mathrm{output}}{\mathrm{total}\mathrm{heat}\mathrm{input}}\times 100\%\\ & =& \frac{350}{900}\times 100\%\\ & =& 40\%\end{array}$

## Factors Affecting the Efficiency of Heat Transfer

In different applications, we may want to design devices to improve or restrict the efficiency of heat transfer in a system depending on the objective. Below we describe some applications that improve heat transfer and some others that restrict heat transfer.

### Improving Heat Transfer Efficiency

#### Cooling fins

Cooling fins use conduction and convection to transfer heat away from objects that generate heat and need to remain cool. They increase the surface area over which heat can be transferred to the surrounding fluid via conduction and convection, and are an efficient method of **improving heat transfer**. The image below shows an example of cooling fins on a motorbike, used to keep the engine cool.

#### Copper-based Saucepans

If you're lucky enough to have a high-quality set of saucepans in your kitchen, you may find they have a copper base. This is because the primary method of heat transfer a saucepan uses to transfer heat from the stove into the food is **conduction**. You might already know that copper is an excellent conductor of electricity, and it's also a great conductor of heat! By using copper in the section of the pan that contacts the stove, there is less thermal resistance to the heat transferring from the stove into the pan. This means the pan can warm up faster and is better to cook with.

### Restricting Heat Transfer Efficiency

#### Double-glazing

Double-glazed windows consist of two panes of glass rather than one. The panes are separated by a layer of gas, usually argon, which is a poor conductor of heat. The two panes are also positioned close enough to each other that a convection current cannot form in the gas between them. The double-glazing is intended to prevent conduction or convection from occurring between the inside of the home and the outside. Double-glazing is an efficient method of **restricting heat transfer**.

#### Oven Gloves

When handling hot objects from an oven, we often use oven gloves to protect our hands from being burnt. These work by separating the hot object and your skin using a thick layer of insulating material such as cotton or silicone. These materials increase the thermal resistance to heat transfer by conduction, which decreases the rate that heat energy is transferred through the glove.

## Heat Transfer Efficiency - Key takeaways

- Heat transfer efficiency is the ratio of the useful output heat energy transfer to the total input heat energy transfer.
- The definition of heat transfer efficiency can be written mathematically as follows

$\begin{array}{rcl}\mathrm{heat}\mathrm{transfer}\mathrm{efficiency}& =& \frac{\mathrm{useful}\mathrm{heat}\mathrm{output}}{\mathrm{total}\mathrm{heat}\mathrm{input}}\\ & & \end{array}$

- Heat transfer occurs through three mechanisms; conduction, convection and radiation.
- Conduction is the mechanism of heat transfer between two substances at different temperatures in direct contact with atomic collisions transferring heat between the substances.
- Convection is the mechanism of heat transfer via liquid and gases which expand when heated.
- Radiation is the energy emitted in the form of infrared electromagnetic radiation by heated surfaces.