Lesson 1 Objective

At the end of this section, you should be able to:

• convert temperature in degree Celsius scale to temperature in Fahrenheit
  scale and vice versa;
• convert temperature in degree Celsius scale to temperature in Kelvin scale
  and vice versa;
• convert temperature in degree Fahrenheit scale to temperature in Kelvin
  scale and vice versa.
• define thermal expansion;
• describe the difference in expansion rate of materials;
• calculate temperature dependent linear expansion of materials.

Brainstorming Question

What happens to metals when they are heated or cooled? Does the wood expand or contract when it is heated or cooled?

key terms and concepts

• Temperature Conversion: Understanding how to convert between Celsius, Fahrenheit, and Kelvin scales is crucial in many scientific fields.
• Celsius to Fahrenheit: Use the formula $T_F$ = $\frac{9}{5}T_C + 32$ for conversion
• Celsius to Kelvin: Add 273.15 to Celsius to get Kelvin; subtract 273.15 from Kelvin to get Celsius.
• Thermal Expansion: Materials expand when heated; the expansion depends on the material’s coefficient of linear expansion and the temperature change.

Conversion between Temperature Scales

Understanding temperature conversion between different scales is essential for many scientific and engineering applications.

Figure 7.1 Temperature conversion

Here’s how you can convert between Celsius, Fahrenheit, and Kelvin scales.

1. Celsius to Fahrenheit Conversion: The formula to convert from Celsius to Fahrenheit is: $T_F$ = $\frac{9}{5}T_C + 32$

Example: If the surrounding temperature is 50°F, the temperature in degrees Celsius is calculated as:

$T_C$ = $\frac{5}{9}(50 – 32)$ = $\frac{5}{9} \times 18$ = $10^\circ C$

2. From Fahrenheit to Celsius: $T_C$ = $\frac{5}{9}(T_F – 32)$

Example: If the temperature of a room is 20°C, the equivalent temperature in degrees Fahrenheit is:

$T_F$ = $\frac{9}{5} \times 20 + 32$ = $36 + 32$ = $68^\circ F$

3. Celsius to Kelvin Conversion:

The relation between Celsius and Kelvin scales is straightforward: $T_K$ = $T_C$ + 273.15

To convert Kelvin to Celsius: $T_C$ = $T_K$ – 273.15

Example: If water is boiled to a temperature of 72°C, it converts to Kelvin as: $T_K$ = 72 + 273.15 = 345.15k

Example: A hot metal at 573.15 K has a temperature in Celsius: $T_C$ = 573.15 -273.15= $300^0\Circ C$

Thermal Expansion of Materials

When materials are heated, they expand. The increase in the length of a solid due to a temperature change is termed linear expansion.

Figure 7.2 When the temperature of the rod raises by ∆T , the length of the rod increases by ∆L.

The equation governing linear expansion is: $\Delta L$ = $\alpha L_0 \Delta T$

where:

• ΔL is the change in length,
• α is the coefficient of linear expansion,
• $L_0$ is the original length, and
• ΔT is the temperature change.

Example: A thin gold rod with an initial length of $1.5 \times 10^{-1} m$ at 27°C falls into a hot sink at 49°C. The change in length is calculated as:

$\Delta L$ = $\alpha L_0 \Delta T$ = $1.4 \times 10^{-5} \, (^\circ C)^{-1} \times 1.5 \times 10^{-1} \, \text{m} \times (49^\circ C – 27^\circ C)$

$\Delta L$ = $\alpha L_0 \Delta T = 1.4 \times 10^{-5} \, (^\circ C)^{-1} \times 1.5 \times 10^{-1} \, \text{m} \times (49^\circ C – 27^\circ C)$

$\Delta L $= $1.4 \times 1.5 \times 22 \times 10^{-6} \, \text{m} = 4.62 \times 10^{-5} \, \text{m}$

Example: A brass rod with an initial length of 50 cm at 25°C is heated to 70°C. The change in length and final length are:

$\Delta L$ = $\alpha L_0 \Delta T$ = $1.9 \times 10^{-5} \, (^\circ C)^{-1} \times 0.5 \, \text{m} \times (70^\circ C – 25^\circ C)$

$\Delta L$ = $1.9 \times 0.5 \times 45 \times 10^{-5} \, \text{m} = 0.043 \, \text{cm}$

$L_F$ = $L_0 + \Delta L$ = $50 \, \text{cm} + 0.043 \, \text{cm}$ = $50.043 \, \text{cm}$