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?
1 answer:
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
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Answer: Option (d) is correct.
Explanation:
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Hence, from the given options the closest answer is of option (d). So, option (d) is correct.