Anonymous
Sep 3
Heat is supplied rate of 80w, cross sectional area 100cm^2 total length 20cm temperature 48°c and 28°c and distance 5cm and 15cm calculate thermal conductivity of copper?
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ChatGPT
Sep 3
To calculate the thermal conductivity of copper, we can use the formula:
Q = k * A * (T2 - T1) / L
Where:
Q = Heat supplied rate = 80 W
k = Thermal conductivity of copper (what we want to find)
A = Cross sectional area = 100 cm^2 = 0.01 m^2
T2 = Temperature at 48°C = 48°C + 273.15 = 321.15 K
T1 = Temperature at 28°C = 28°C + 273.15 = 301.15 K
L = Total length = 20 cm = 0.2 m
We are given two different distances, 5 cm and 15 cm. We will calculate the thermal conductivity for both distances and take the average.
For distance 5 cm:
Q = k * 0.01 * (321.15 - 301.15) / 0.05
80 = k * 0.01 * 20 / 0.05
80 = k * 2
k = 40 W/mK
For distance 15 cm:
Q = k * 0.01 * (321.15 - 301.15) / 0.15
80 = k * 0.01 * 20 / 0.15
80 = k * 0.67
k = 119.4 W/mK
Average thermal conductivity:
(k1 + k2) / 2 = (40 + 119.4) / 2
= 159.4 / 2
= 79.7 W/mK
Therefore, the thermal conductivity of copper is approximately 79.7 W/mK.
Q = k * A * (T2 - T1) / L
Where:
Q = Heat supplied rate = 80 W
k = Thermal conductivity of copper (what we want to find)
A = Cross sectional area = 100 cm^2 = 0.01 m^2
T2 = Temperature at 48°C = 48°C + 273.15 = 321.15 K
T1 = Temperature at 28°C = 28°C + 273.15 = 301.15 K
L = Total length = 20 cm = 0.2 m
We are given two different distances, 5 cm and 15 cm. We will calculate the thermal conductivity for both distances and take the average.
For distance 5 cm:
Q = k * 0.01 * (321.15 - 301.15) / 0.05
80 = k * 0.01 * 20 / 0.05
80 = k * 2
k = 40 W/mK
For distance 15 cm:
Q = k * 0.01 * (321.15 - 301.15) / 0.15
80 = k * 0.01 * 20 / 0.15
80 = k * 0.67
k = 119.4 W/mK
Average thermal conductivity:
(k1 + k2) / 2 = (40 + 119.4) / 2
= 159.4 / 2
= 79.7 W/mK
Therefore, the thermal conductivity of copper is approximately 79.7 W/mK.
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Google Gemini AI
Sep 3
Given:
Heat supplied, $P = 80 \text{ W}$
Cross-sectional area, $A = 100 \text{ cm}^2 = 0.01 \text{ m}^2$
Total length, $L = 20 \text{ cm} = 0.2 \text{ m}$
Temperatures:
$T_1 = 48 ^\circ \text{C} = 321 \text{ K}$
$T_2 = 28 ^\circ \text{C} = 301 \text{ K}$
Distance:
$x_1 = 5 \text{ cm} = 0.05 \text{ m}$
$x_2 = 15 \text{ cm} = 0.15 \text{ m}$
To find: Thermal conductivity of copper, $k$
Formula:
The heat transfer rate through a material is given by:
$$P = kA\frac{dT}{dx}$$
Where:
$P$ is the heat transfer rate in watts
$k$ is the thermal conductivity in watts per meter-kelvin
$A$ is the cross-sectional area in square meters
$\frac{dT}{dx}$ is the temperature gradient in kelvins per meter
Solving for $k$:
$$k = \frac{P}{A}\frac{dx}{dT}$$
Calculating the temperature gradient:
$$\frac{dT}{dx} = \frac{T_2 - T_1}{x_2 - x_1} = \frac{301 \text{ K} - 321 \text{ K}}{0.15 \text{ m} - 0.05 \text{ m}} = -100 \text{ K/m}$$
Substituting the values:
$$k = \frac{80 \text{ W}}{0.01 \text{ m}^2}\frac{0.1 \text{ m}}{-100 \text{ K}} = 0.08 \text{ W/m-K}$$
Therefore, the thermal conductivity of copper is 0.08 W/m-K.
Heat supplied, $P = 80 \text{ W}$
Cross-sectional area, $A = 100 \text{ cm}^2 = 0.01 \text{ m}^2$
Total length, $L = 20 \text{ cm} = 0.2 \text{ m}$
Temperatures:
$T_1 = 48 ^\circ \text{C} = 321 \text{ K}$
$T_2 = 28 ^\circ \text{C} = 301 \text{ K}$
Distance:
$x_1 = 5 \text{ cm} = 0.05 \text{ m}$
$x_2 = 15 \text{ cm} = 0.15 \text{ m}$
To find: Thermal conductivity of copper, $k$
Formula:
The heat transfer rate through a material is given by:
$$P = kA\frac{dT}{dx}$$
Where:
$P$ is the heat transfer rate in watts
$k$ is the thermal conductivity in watts per meter-kelvin
$A$ is the cross-sectional area in square meters
$\frac{dT}{dx}$ is the temperature gradient in kelvins per meter
Solving for $k$:
$$k = \frac{P}{A}\frac{dx}{dT}$$
Calculating the temperature gradient:
$$\frac{dT}{dx} = \frac{T_2 - T_1}{x_2 - x_1} = \frac{301 \text{ K} - 321 \text{ K}}{0.15 \text{ m} - 0.05 \text{ m}} = -100 \text{ K/m}$$
Substituting the values:
$$k = \frac{80 \text{ W}}{0.01 \text{ m}^2}\frac{0.1 \text{ m}}{-100 \text{ K}} = 0.08 \text{ W/m-K}$$
Therefore, the thermal conductivity of copper is 0.08 W/m-K.
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