Topic: Heat Transfer
An electric heater is sandwiched between two plates each 0.3 m long and 0.1 m wide with a thickness of 30 mm. At steady state condition the heater is maintained at a temperature of $100^{\circ}C.$ with a current of 0.25 A and voltage of 200 V. Assume the plates are perfectly insulated at the edges, and the heater is having perfect contact with the plates to given a temperature of $50^{\circ}C$ on the outside of the plate surface. What is the thermal conductivity of the plate material?
- A. $1.0~W/m-K$
- B. $0.5~W/m-K$
- C. $0.015~W/m-K$
- D. $0.3~W/m-K$
Answer: B. $0.5~W/m-K$
Explanation: Total heat generated $H_{gen} = V \times I = 200 \times 0.25 = 50~W$. Since heat flows equally through both plates, heat through one plate is $Q = 50/2 = 25~W$. From Fourier's law, $Q = kA\frac{\Delta T}{\Delta x}$. Substituting values: $25 = k \times (0.3 \times 0.1) \times \frac{100 - 50}{0.030}$. $25 = k \times 0.03 \times \frac{50}{0.03} \Rightarrow 25 = 50k \Rightarrow k = 0.5~W/m-K$.