Question

An electric kitchen range has a total wall area of 1.40 m2 and is insulated with a layer of fiberglass 4.0 cm thick. The inside surface of the fiberglass has a temperature of 175 ∘C and its outside surface is at 35 ∘C. The fiberglass has a thermal conductivity of 0.040 W/(m⋅K).What is the heat current through the insulation, assuming it may be treated as a flat slab with an area of 1.40 m2?What electric-power input to the heating element is required to maintain this temperature?

Answer 1

Answer: Heat current through the insulator=196W

Electric power= 196W

Explanation: Given: Kglass = 0.040W/m

Temperature of inside glassTi=175°C

Temperature of outside glass To= 35°C

Area=1.4m^2 , L= 4×10^-2

Heat current(H)= K ×A× (Ti - To)/L

Substituting the values into the equation

H = 0.04 × K × 1.4 ×(175-35)/4×10^-2

H= 196W.

The electric power = Heat current =196W

The electric power is the magnitude of heat current