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?

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

i believe the answer is electric potential.

explanation:

a cold ice-tea.

pour in room temperature or cold water. cover the pitcher and refrigerate. steep white or green tea for 6 to 8 hours; steep black or oolong tea for 8 to 12 hours.jun 28, 2015