A copper rod of length 50 cm and diameter 4.0 centimetre is given.If the termed conductivity of Copper is 4.2 * 10 to the power 2 WM -1K -1 calculate its thermal resistivity .
Answers
Answer:
Thermal resistivity of the copper rod is 2.33 × 10-³ K m W-1, and the rate of flow of heat through it is 34.759 W.
i) Given, thermal conductivity of copper (k) = 4.2 × 102 W m⁻¹ K⁻¹
=> Resistivity (R) = 1/conductivity (k)
=> R = 1/ (4.2 × 102 W m⁻¹ K⁻¹ )
=> R = 1 / 428.4 W m⁻¹ K⁻¹
=> R = 0.00233 K m W⁻¹
=> R = 2.33 × 10⁻³ K m W⁻¹
ii) Also, k = (QL) / (A∆T) -(i)
Where Q is the rate of heat flowing through the rod,
L is the length of the rod,
A is the area of cross-section of the rod,
∆T is the temperautre difference between the ends of the rod.
• Given,
k = 4.2 × 102 W m⁻¹ K⁻¹,
L = 50 cm = 50 / 100 m = 0.5 m,
Diameter (D) = 4 cm
=> R = D/2
=> R = 2 cm = 2/100 m
=> R = 0.02 m
Therefore, area (A) = πr² = 3.14 × (0.02 m)²
=> A = 0.001256 m² = 1.256 × 10⁻³ m²
∆T = 50°C = 50 + 273 K
=> ∆T = 323 K
Q = ?
• Substituting these values in equation (i), we get,
4.2 × 102 W m⁻¹ K⁻¹ = (Q × 0.5 m) / (1.256 × 10⁻³ m² × 323 K)
=> 428.4 W m⁻¹ K⁻¹ = (Q × 0.5 m) / (405.68 × 10⁻³ m²K)
=> Q = (428.4 W m⁻¹ K⁻¹ × 405.68 × 10⁻³ m²K) / 0.5 m
=> Q = 34759.34 × 10⁻³ W
=> Q = 34.759 × 10³ × 10⁻³ W
=> Q = 34.759 W