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\( 2^{\text {ND }} \) ASSIGNMENI Subject: Applied Physics-1 ( 2024-2025), Semester I NAME : REGISTRATION NUMBER : BRANCH : Choose correct answer (put \( \checkmark \) ) of the following questions. ( \( 1 \times 10 \) ) 1 A copper rod of length 1 m and cross-sectional area \( 0.005 \mathrm{~m}^{\mathbf{2}} \) has its ends maintained at \( 100^{\circ} \mathrm{C} \) and \( 0^{\circ} \mathrm{C} \). If the thermal conductivity of copper is \( \mathbf{4 0 0} \mathrm{W} / \mathrm{m} \cdot \mathrm{K} \). The rate of heat transfer is (a) 200 Watt (b) 1470 watt (c) 1050 Watt (d) 1260 Watt .

Ask by Rodriquez Cole. in India
Feb 27,2025

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Answer

The rate of heat transfer is 200 Watt.

Solution

We use Fourier’s law for heat conduction, which states that the rate of heat transfer (P) is given by:   P = (k × A × ΔT) / L where:   k = thermal conductivity = 400 W/m·K   A = cross-sectional area = 0.005 m²   ΔT = temperature difference = (100°C – 0°C) = 100 K   L = length of the rod = 1 m Substitute the given values:   P = (400 × 0.005 × 100) / 1     = (400 × 0.5)     = 200 Watt Thus, the correct answer is option (a) 200 Watt.

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To determine the rate of heat transfer through the copper rod, you can use Fourier's law of heat conduction, which is given by the formula: \[ Q = \frac{k \cdot A \cdot (T_1 - T_2)}{L} \] where \( Q \) is the rate of heat transfer in watts (W), \( k \) is the thermal conductivity, \( A \) is the cross-sectional area, \( T_1 \) and \( T_2 \) are the temperatures at the ends, and \( L \) is the length of the rod. Given: - \( k = 400 \, \mathrm{W/(m \cdot K)} \) - \( A = 0.005 \, \mathrm{m^2} \) - \( T_1 = 100 \, \mathrm{°C} = 373.15 \, \mathrm{K} \) - \( T_2 = 0 \, \mathrm{°C} = 273.15 \, \mathrm{K} \) - \( L = 1 \, \mathrm{m} \) Calculating the temperature difference: \[ T_1 - T_2 = 100 - 0 = 100 \, \mathrm{K} \] Substituting into the formula: \[ Q = \frac{400 \, \mathrm{W/(m \cdot K)} \cdot 0.005 \, \mathrm{m^2} \cdot 100 \, \mathrm{K}}{1 \, \mathrm{m}} = 200 \, \mathrm{W} \] Thus, the correct answer is (a) 200 Watt.

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