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?

Respuesta :

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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The heat current through the insulation, treating it as a flat slab with an area of 1.40 m² is 196 W and the electric power input to the heating element required to maintain the temperature in the kitchen is also 196 W.

Given to us,

Area of the flat slab, A = 1.40 m²,

the thickness of the fiberglass, d = 4 cm = 0.04 m,

Inside surface of the fiberglass, [tex]T_h[/tex] = 175°C = 175°C + 273.15 = 448.15 K,

Outside surface of the fiberglass, [tex]T_i[/tex] = 35°C = 35°C + 273.15 = 308.15 K,

thermal conductivity of fiberglass, K = 0.040 W/(mK),

a.)

Conduction is the process by which heat energy is been transferred due to the collisions between atoms and molecules in solids. The heat transfer in conduction is written as,

[tex]\bold{\dfrac{\dot Q}{t} = \dfrac{K A (T_h-T_i)}{d}}[/tex]

[tex]{\dfrac{\dot Q}{t} = \dfrac{0.040\times 1.40 (448.15-308.15)}{0.04}}[/tex]

[tex]{\dfrac{\dot Q}{t} = 196\ W[/tex]

Therefore, the heat current through the insulation, treating it as a flat slab with an area of 1.40 m² is 196 W.

b.)

As the electric-power input to the heating element required to maintain the temperature in the kitchen is 196 W, because as the fiberglass is constantly losing heat, therefore to maintain the temperature we need to continuously provide that amount of heat to the kitchen for which the same amount of work will be required.

Thus, the electric-power input to the heating element required to maintain the temperature in the kitchen is 196 W.

Hence, the heat current through the insulation, treating it as a flat slab with an area of 1.40 m² is 196 W and the electric power input to the heating element required to maintain the temperature in the kitchen is also 196 W.

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