Answer:
A. T=15.54 °C
B. Q/A= 0.119 W/m2
Step-by-step explanation:
To solve this problem we need to use the Fourier's law for thermal conduction:
![Q= kA(dT)/(dx)](https://img.qammunity.org/2020/formulas/physics/college/ylx0euehl5f4yiwwlja2cb1def9h95hsf1.png)
Here, the rate of flow per square meter must be the same through the complete wall. Therefore, we can use it to find the temperature at the plane where the wood meets the Styrofoam as follows:
![(Q)/(A) =(T_1-T_0)/(d_w)k_w=(T_2-T_1)/(d_s)k_s\\T_1((k_w)/(d_w)+(k_s)/(d_s))=T_2(k_s)/(d_s)+T_0(k_w)/(d_w)\\T_1=(T_2(k_s)/(d_s)+T_0(k_w)/(d_w))/((k_w)/(d_w)+(k_s)/(d_s))\\T_1= 15.54 \°C](https://img.qammunity.org/2020/formulas/physics/college/vnp28i3xrwa1o2jzjvq51gyey9z13v7t2t.png)
Then, to find the rate of heat flow per square meter, we have:
![(Q)/(A)=(T_1-T_0)/(d_w)k_w=0.119 (W)/(m^2)\\(Q)/(A)=(T_2-T_1)/(d_s)k_s= 0.119 (W)/(m^2)](https://img.qammunity.org/2020/formulas/physics/college/d5fl8wnhsbpd651fjg5dntr77ehofa96zp.png)
![T_0: Temperature \ in \ the \ house\\T_1: Temperature \ at \ the \ plane \ between \ wood \ and \ styrofoam\\T_2: Temperature \ outside\\k_w: k \ for \ wood\\d_w: wood \ thickness\\k_s: k \ for \ styrofoam\\d_s: styrofoam \ thickness](https://img.qammunity.org/2020/formulas/physics/college/90eu4mik6qu4ahjqzb426dbotpbk7e9jde.png)