Heat, Temperature & Thermal Conductivity
Heat and Temperature
Temperature is the measure of how hot or cold a body is. It is related to the average kinetic energy of the molecules in a substance.
Common temperature scales are Celsius (°C) and Kelvin (K). The SI unit is Kelvin.
Conversion:
[
K = °C + 273
]
Heat is energy transferred from a body at higher temperature to one at lower temperature due to temperature difference. Heat is measured in Joules (J).
Heat is not stored in a body; it is only transferred.
Thermal Conductivity
Thermal conductivity is a property of a material that shows how easily heat flows through it.
Metals → high conductivity
Wood, brick, foam → low conductivity
Fourier’s Law
\dot{Q} = -kA\frac{dT}{dx}
For steady heat flow through a slab:
\dot{Q} = kA\frac{\Delta T}{L}
Where:
( \dot{Q} ) = heat transfer rate (W)
( k ) = thermal conductivity
( A ) = area
( \Delta T ) = temperature difference
( L ) = thickness
Example Question
A glass window is 0.8 cm thick with an area of 1.5 m². The inside temperature is 22°C while the outside temperature is 2°C. If the thermal conductivity of glass is 0.84 W/m·K, calculate the rate of heat loss through the window.
Solution
Convert thickness:
[
L = 0.8 , cm = 0.008 , m
]
Temperature difference:
[
\Delta T = 22 - 2 = 20K
]
Area:
[
A = 1.5 , m^2
]
Apply Fourier’s law:
[
\dot{Q} = kA\frac{\Delta T}{L}
]
Substitute values:
[
\dot{Q} = 0.84 \times 1.5 \times \frac{20}{0.008}
]
[
\dot{Q} = 0.84 \times 1.5 \times 2500
]
[
\dot{Q} = 0.84 \times 3750
]
[
\dot{Q} = 3150 , W
]
Final Answer
Heat loss through the window is 3150 W.
Example Question
A metal rod of length 0.5 m and cross-sectional area 0.02 m² transfers 400 J of heat in 10 seconds. The ends are maintained at 100°C and 20°C. Calculate the thermal conductivity of the metal.
Solution
Rate of heat transfer:
[
\dot{Q} = \frac{400}{10} = 40 , W
]
Temperature difference:
[
\Delta T = 100 - 20 = 80K
]
Apply Fourier’s law:
[
\dot{Q} = kA\frac{\Delta T}{L}
]
Rearrange:
[
k = \frac{\dot{Q}L}{A\Delta T}
]
Substitute values:
[
k = \frac{40 \times 0.5}{0.02 \times 80}
]
[
k = \frac{20}{1.6}
]
[
k = 12.5 , W/m·K
]
Final Answer
Thermal conductivity of the metal is 12.5 W/m·K.
Example Question
A concrete wall has thermal conductivity 1.1 W/m·K, thickness 0.15 m, and area 20 m². The temperature difference across the wall is 15°C. Calculate the heat lost in 1 hour.
Solution
Convert time:
[
t = 1 , hour = 3600 , s
]
Apply Fourier’s law:
[
\dot{Q} = kA\frac{\Delta T}{L}
]
Substitute:
[
\dot{Q} = 1.1 \times 20 \times \frac{15}{0.15}
]
[
\dot{Q} = 1.1 \times 20 \times 100
]
[
\dot{Q} = 2200 , W
]
Total heat in 1 hour:
[
Q = \dot{Q} \times t
]
[
Q = 2200 \times 3600
]
[
Q = 7.92 \times 10^6 , J
]
Final Answer
Heat lost in 1 hour = 7.92 × 10⁶ J
Example Question
A wooden door has thermal conductivity 0.12 W/m·K, area 2 m², and temperature difference 25°C. If the maximum heat loss allowed is 50 W, determine the required thickness.
Solution
Apply Fourier’s law:
[
\dot{Q} = kA\frac{\Delta T}{L}
]
Rearrange:
[
L = \frac{kA\Delta T}{\dot{Q}}
]
Substitute:
[
L = \frac{0.12 \times 2 \times 25}{50}
]
[
L = \frac{6}{50}
]
[
L = 0.12 , m
]
Final Answer
Required thickness = 0.12 m