Consider two rods 1 and 2 of same length. They have different specific heats (C₁, C₂) , thermal conductivities (K₁, K₂) and area of cross-section (A₁, A₂) respectively. Both the rods have temperatures (T₁, T₂) at their ends. If their rate of loss of heat due to conduction is equal, then

Step-by-step solution

The rate of heat flow (or rate of loss of heat due to conduction) through a rod is given by Fourier's law of heat conduction: dQ dt = KA(T₁ - T₂) L For the first rod, the rate of heat flow is: ( dQ dt )₁ = K₁ A₁ (T₁ - T₂) L For the second rod, the rate of heat flow is: ( dQ dt )₂ = K₂ A₂ (T₁ - T₂) L It is given that the rate of loss of heat due to conduction is equal for both rods. Equating the two rates: K₁ A₁ (T₁ - T₂) L = K₂ A₂ (T₁ - T₂) L Since the length L and the temperature difference (T₁ - T₂) are the same