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Thermal conductivity of a material varies linearly with temperature such that k = 20 W/m-°C at 100°C&k = 10 W/m-°C at 40°C. A plane slab 10 cm thick of same material has it’s left and right surfaces maintained a temperatures 20°C & 80°C. What is the rate of heat conduction through the slab in W/m2 ?
- a)7.0
- b)7.0
Correct answer is between '7.0,7.0'. Can you explain this answer?
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Thermal conductivity of a material varies linearly with temperature s...
Given
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k = aT + b
k = 20W/m-°C at T = 40°C
k = 10W/m-°C at T = 40°C
Substituting the given conditions
20 = 100a = b
10 = 40a = b
∴
∴
For the slab
But as is a linear function of T
Where kavg is k for Tavg = 
Thus in this case
Kavg = k50
Or 7 kW/m2
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Thermal conductivity of a material varies linearly with temperature s...
Understanding Thermal Conductivity Variation
The thermal conductivity (k) of the material varies linearly with temperature. Given two specific points:
- k = 20 W/m-°C at 100°C
- k = 10 W/m-°C at 40°C
To find the thermal conductivity at intermediate temperatures, we derive the linear relationship.
Calculating the Linear Equation
1. Determine the slope (m):
m = (k2 - k1) / (T2 - T1) = (20 - 10) / (100 - 40) = 10 / 60 = 1/6 W/m-°C².
2. Form the equation:
k = (1/6)(T - 40) + 10.
Finding k at Given Temperatures
We need to find k at T1 = 20°C and T2 = 80°C.
- For T1 = 20°C:
k(20) = (1/6)(20 - 40) + 10 = -1.67 + 10 = 8.33 W/m-°C.
- For T2 = 80°C:
k(80) = (1/6)(80 - 40) + 10 = 6.67 + 10 = 16.67 W/m-°C.
Calculating Average Thermal Conductivity
The average thermal conductivity (k_avg) between the two temperatures can be calculated:
k_avg = (k(20) + k(80)) / 2 = (8.33 + 16.67) / 2 = 12.5 W/m-°C.
Applying Fourier's Law of Heat Conduction
Using Fourier's Law, the rate of heat conduction (q) through the slab can be expressed as:
q = k_avg * (ΔT / L),
Where:
- ΔT = T_right - T_left = 80°C - 20°C = 60°C,
- L = thickness of the slab = 0.1 m.
Thus,
q = 12.5 W/m-°C * (60 / 0.1) = 12.5 * 600 = 7500 W/m².
However, since the question suggests a value between '7.0, 7.0', the calculation must have a specific context or additional constraints not provided.
The thermal conductivity (k) of the material varies linearly with temperature. Given two specific points:
- k = 20 W/m-°C at 100°C
- k = 10 W/m-°C at 40°C
To find the thermal conductivity at intermediate temperatures, we derive the linear relationship.
Calculating the Linear Equation
1. Determine the slope (m):
m = (k2 - k1) / (T2 - T1) = (20 - 10) / (100 - 40) = 10 / 60 = 1/6 W/m-°C².
2. Form the equation:
k = (1/6)(T - 40) + 10.
Finding k at Given Temperatures
We need to find k at T1 = 20°C and T2 = 80°C.
- For T1 = 20°C:
k(20) = (1/6)(20 - 40) + 10 = -1.67 + 10 = 8.33 W/m-°C.
- For T2 = 80°C:
k(80) = (1/6)(80 - 40) + 10 = 6.67 + 10 = 16.67 W/m-°C.
Calculating Average Thermal Conductivity
The average thermal conductivity (k_avg) between the two temperatures can be calculated:
k_avg = (k(20) + k(80)) / 2 = (8.33 + 16.67) / 2 = 12.5 W/m-°C.
Applying Fourier's Law of Heat Conduction
Using Fourier's Law, the rate of heat conduction (q) through the slab can be expressed as:
q = k_avg * (ΔT / L),
Where:
- ΔT = T_right - T_left = 80°C - 20°C = 60°C,
- L = thickness of the slab = 0.1 m.
Thus,
q = 12.5 W/m-°C * (60 / 0.1) = 12.5 * 600 = 7500 W/m².
However, since the question suggests a value between '7.0, 7.0', the calculation must have a specific context or additional constraints not provided.
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Thermal conductivity of a material varies linearly with temperature such that k = 20 W/m-°C at 100°C&k = 10 W/m-°C at 40°C. A plane slab 10 cm thick of same material has it’s left and right surfaces maintained a temperatures 20°C & 80°C. What is the rate of heat conduction through the slab in W/m2 ?a)7.0b)7.0Correct answer is between '7.0,7.0'. Can you explain this answer?
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