Consider the setup depicted in the figure. Assume that the thermal conductivity of the bent part of the frame is 780 J s ⁻¹ m ⁻¹ ^ C ⁻¹ , while for the straight part it is 390 J s ⁻¹ m ⁻¹ ^ C ⁻¹ . Determine the ratio of the rate of heat flow through the bent part to the rate of heat flow through the straight part.

Step-by-step solution

Let the temperature difference across the parallel section be T . The rate of heat flow through a conductor is given by H = KA T L , where K is the thermal conductivity, A is the cross-sectional area, and L is the length. For the bent part: Thermal conductivity, K_b = 780 J s ⁻¹ m ⁻¹ ^ C ⁻¹ Length of the bent part, L_b = 5 cm + 60 cm + 5 cm = 70 cm Rate of heat flow, H_b = K_b A T L_b For the straight part in parallel: Thermal conductivity, K_s = 390 J s ⁻¹ m ⁻¹ ^ C ⁻¹ Length of the straight part, L_s = 60 cm Rate