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Heat Transfer - Conductors and Insulators

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Conduction is the process of heat transfer through a substance without the bulk movement of the material itself. It primarily occurs in solids through particle collisions.

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Thermal Conductors are materials that allow heat to flow through them easily. Metals are excellent conductors because they possess free electrons that can move rapidly, transferring kinetic energy. Examples include Copper (CuCu) and Aluminium (AlAl).

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Thermal Insulators are materials that do not allow heat to flow through them easily. They lack free electrons and have fixed molecular structures. Examples include wood, glass, plastic, and air.

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The rate of heat transfer through a material depends on its thermal conductivity constant (kk), the surface area (AA), the temperature difference (ΔT\Delta T), and the thickness of the material (dd).

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Real-world application: Double-glazed windows use a layer of trapped air (an insulator) between two glass panes to minimize heat loss via conduction, as air has a very low kk value.

📐Formulae

Q=k⋅A⋅ΔT⋅tdQ = \frac{k \cdot A \cdot \Delta T \cdot t}{d}

P=Qt=k⋅A⋅ΔTdP = \frac{Q}{t} = \frac{k \cdot A \cdot \Delta T}{d}

ΔT=Thot−Tcold\Delta T = T_{hot} - T_{cold}

💡Examples

Problem 1:

A glass window pane has an area of 2.02.0 m2m^2 and a thickness of 0.0050.005 mm. The temperature inside the room is 2020 ∘C^{\circ}C and the temperature outside is 55 ∘C^{\circ}C. If the thermal conductivity of glass is 0.80.8 W⋅m−1⋅K−1W \cdot m^{-1} \cdot K^{-1}, calculate the rate of heat loss PP through the window.

Solution:

Given: k=0.8k = 0.8 W⋅m−1⋅K−1W \cdot m^{-1} \cdot K^{-1} A=2.0A = 2.0 m2m^2 ΔT=20−5=15\Delta T = 20 - 5 = 15 KK (or ∘C^{\circ}C) d=0.005d = 0.005 mm

Using the formula: P=k⋅A⋅ΔTdP = \frac{k \cdot A \cdot \Delta T}{d} P=0.8×2.0×150.005P = \frac{0.8 \times 2.0 \times 15}{0.005} P=240.005P = \frac{24}{0.005} P=4800P = 4800 WW

Explanation:

The rate of heat transfer PP is calculated by multiplying the thermal conductivity, area, and temperature difference, then dividing by the thickness. The unit of the result is Watts (WW), representing Joules per second.

Problem 2:

Calculate the temperature difference ΔT\Delta T required to transfer 10001000 JJ of heat through a metal plate of area 0.10.1 m2m^2 and thickness 0.020.02 mm in 1010 seconds, given k=50k = 50 W⋅m−1⋅K−1W \cdot m^{-1} \cdot K^{-1}.

Solution:

First, find the power (rate of heat transfer): P=Qt=100010=100P = \frac{Q}{t} = \frac{1000}{10} = 100 WW

Rearrange the formula to solve for ΔT\Delta T: ΔT=P⋅dk⋅A\Delta T = \frac{P \cdot d}{k \cdot A} ΔT=100×0.0250×0.1\Delta T = \frac{100 \times 0.02}{50 \times 0.1} ΔT=25\Delta T = \frac{2}{5} ΔT=0.4\Delta T = 0.4 KK

Explanation:

By rearranging the conduction equation, we can find the required temperature gradient. A higher thermal conductivity kk means a smaller temperature difference is needed to move the same amount of heat.