# Solution Let's break down the problem step by step. ## Given Data - **Area of each solid, \( A \) = .2 m²** - **Thickness of each solid, \( L \) = 50 mm = .05 m** - **Thermal conductivity of A, \( k_A \) = 32 W/mK** - **Thermal conductivity of B, \( k_B \) = 10 W/mK** - **Power supplied by heater, \( Q = 300 \) W** - **Surface temperature on both sides, \( T_1 = T_3 = 310 \) K** - **Perfect thermal contact at interfaces** Diagram: ``` T1 (310K) --| Solid A |-- T2 --| Heater |-- T3 --| Solid B |-- (310K) ``` --- ## a) **Calculate the heat flux through each solid** The heat supplied by the heater must flow through both solids A and B in series, so the heat flux through both is the same. \[ Q = \dot{q} \cdot A \] \[ \dot{q} = \frac{Q}{A} = \frac{300}{.2} = 150\, \text{W/m}^2 \] **Answer:** **Heat flux through each solid = 150 W/m²** --- ## b) **What is the operating temperature of the heater?** Let's denote: - \( T_1 = 310\,K \) (left external surface) - \( T_2 \) (interface between A and heater) - \( T_3 \) (interface between B and heater) - \( T_h \) (temperature of the heater, assumed uniform) The heat transfer through each solid: - Through A: \[ Q = \frac{k_A A (T_2 - T_1)}{L} \] - Through B: \[ Q = \frac{k_B A (T_3 - T_4)}{L} \] Where \( T_4 = 310\,K \) (right external surface). The heater is sandwiched between A and B, so the temperature at the heater surface is \( T_h \), and: - \( T_2 = T_h \) (interface with A) - \( T_3 = T_h \) (interface with B) So: - \( Q = \frac{k_A A (T_h - T_1)}{L} \) - \( Q = \frac{k_B A (T_h - T_4)}{L} \) Both sides see the same \( Q \), so: - On the left: \[ T_h - T_1 = \frac{Q L}{k_A A} \] - On the right: \[ T_h - T_4 = \frac{Q L}{k_B A} \] Plug in the numbers: - For A: \[ T_h - 310 = \frac{300 \times .05}{32 \times .2} = \frac{15}{6.4} = 2.34375\,K \] \[ T_h = 310 + 2.34375 = 312.34\,K \] - For B: \[ T_h - 310 = \frac{300 \times .05}{10 \times .2} = \frac{15}{2} = 7.5\,K \] \[ T_h = 310 + 7.5 = 317.5\,K \] **But the heater is in the middle, and the temperature must be such that the heat flows are the same; so we need to solve for \( T_h \) considering both sides:** Let \( Q = \frac{k_A A (T_h - 310)}{L} \) (left) and \( Q = \frac{k_B A (T_h - 310)}{L} \) (right). However, both solids are exposed to 310 K at their external faces, and the heater sits between, so the heat supplied by the heater drives flow in both directions, and the temperature of the heater should be higher than 310 K. But since the heater is sandwiched by two different thermal resistances, the temperature rise above 310 K is the sum of the temperature rises across each slab. \[ Q = \frac{T_h - 310}{R_{\text{total}}} \] Where: \[ R_{\text{total}} = R_A + R_B = \frac{L}{k_A A} + \frac{L}{k_B A} \] Calculate each resistance: - \( R_A = \frac{.05}{32 \times .2} = \frac{.05}{6.4} = .0078125 \) K/W - \( R_B = \frac{.05}{10 \times .2} = \frac{.05}{2} = .025 \) K/W So, \[ R_{\text{total}} = .0078125 + .025 = .0328125 \text{ K/W} \] Now, \[ T_h = 310 + Q \times R_{\text{total}} = 310 + 300 \times .0328125 = 310 + 9.84375 = 319.84\,K \] --- ## **Summary of Answers** ### a) **Heat flux through each solid** \[ \boxed{150\, \text{W/m}^2} \] ### b) **Operating temperature of the heater** \[ \boxed{319.84\,K} \]