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A carpenter builds an exterior house wall with a layer of wood 3.0 cm thick on the outside and a layer of Styrofoam insulation 2.4 cm thick on the inside wall surface. The wood has k=0.080W/(m?K), and the Styrofoam has k= 0.010 W/(m?K). The interior surface temperature is 20.0 ?C , and the exterior surface temperature is -13.0 ?C

A.)What is the temperature at the plane where the wood meets the Styrofoam? _______ Celsius

B.)What is the rate of heat flow per square meter through this wall? ______W/m^2

2 Answers

2 votes

Final answer:

The temperature at which the wood meets the Styrofoam is determined using the proportional temperature drops across each material, considering their respective thermal resistances. The rate of heat flow per square meter through the wall is calculated using Fourier's law, considering the combined thermal resistance of both layers.

Step-by-step explanation:

Temperature at the Wood-Styrofoam Plane and Heat Flow Rate

To find the temperature at the plane where the wood meets the Styrofoam, we need to use the concept of thermal resistance and the fact that the heat flow through both materials is the same. With the given thermal conductivity (k values) and thicknesses of the wood and Styrofoam, we can calculate their respective thermal resistances (R = thickness/k). Then, by setting up a proportion, we can find the temperature drop across the wood (ΔTw) and the Styrofoam (ΔTs) using the formula ΔT = (k*A* ΔT)/(thickness), where A is the area (which cancels out as it's the same for both layers). Knowing the temperature drop across each and the outer and inner surface temperatures, we can find the temperature at the plane where the layers meet.

To calculate the rate of heat flow per square meter through this wall, we'll use Fourier's law of thermal conduction, which is given by Q = k*A* ΔT/d, where Q is the heat flow rate, A is the area, ΔT is the temperature difference, and d is the thickness. Since the wall consists of two layers with different thermal conductivities and thicknesses, we need to calculate the equivalent thermal resistance for the combined wall and then use the overall temperature difference to find the heat flow rate.

User Jacquetta
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Answer:

Step-by-step explanation:

Given

thermal conductivity of wood
K_w=0.08 W/m^2-K

thermal conductivity of insulation
K_i=0.01 W/m^2-K

thickness of wood
t_2=3 cm

thickness of insulation
t_1=2.4 cm


T_i=20^(\circ)C


T_o=-13^(\circ)C

we know heat Flow is given by


Q=kA(dT)/(dx)


dT= change in temperature


dx= thickness

K=thermal conductivity

A=Area of cross-section

A is same

Suppose T is the temperature of Junction

as heat Flow is same thus


(k_w(20-T))/(3)=(k_i(T-(-13)))/(2.4)


(0.08(20-T))/(3)=(0.01(T+13))/(2.4)


T=19.36 ^(\circ)C

(b)Rate of heat flow


Q=(k_w(T+13))/(3* 10^(-2))


Q=(0.08* 32.36)/(0.03)


Q=86.303 W/m^2

User Compid
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