Consider two rods of same length and different specific heats $ (S_1, S_2) $ , conductivities $ (K_1, K_2)$ and area of cross-sections $ (A_1 , A_2) $ and both having temperatures $ T_1$ and $T_2 $ at their ends. If rate of loss of heat due to conduction is equal, then :-

Question

If current in a conductor $3t^2 + 4t^3$, charge = $?$, flow $t = 1\; to\; t = 2s$

Answer 1

To solve this problem, we need to understand the formula for the rate of heat conduction through a rod. The formula given by Fourier's law of heat conduction is:

q = \frac{dQ}{dt} = -K A \frac{dT}{dx}

where:

q is the rate of heat transfer (in Watts, if using SI units),
K is the thermal conductivity of the material,
A is the cross-sectional area of the rod,
\frac{dT}{dx} is the temperature gradient along the length of the rod.

For both rods, given that they have the same length and temperature difference (T_2 - T_1), the temperature gradient \frac{dT}{dx} is the same. Therefore, the rate of heat loss per unit time for each rod is:

q_1 = K_1 A_1 \frac{(T_2 - T_1)}{L}

q_2 = K_2 A_2 \frac{(T_2 - T_1)}{L}

According to the problem, both rods have the same rate of loss of heat due to conduction. Therefore, we equate these two expressions:

K_1 A_1 \frac{(T_2 - T_1)}{L} = K_2 A_2 \frac{(T_2 - T_1)}{L}

Since (T_2 - T_1) and L are constant and non-zero for both rods, they can be canceled out from both sides of the equation:

K_1 A_1 = K_2 A_2

Thus, the correct answer is: K_1 A_1 = K_2 A_2

This means that the thermal conduction property (i.e., the product of thermal conductivity and cross-sectional area) must be the same for both rods to achieve the same rate of heat transfer, given the same temperature difference and length.

Answer 2