Answer
$340Btu/h$ to $350Btu/h$
Work Step by Step
$\frac{Q}{t}=kA\frac{T_1-T_2}{l}$
If the temperature inside and outside the house has been constant for some period of time, the rate of heat flow should be constant through both materials.
$T_o$ temperature outside house
$T_i$ temperature inside house
$T_w$ temperature inside wall
$\frac{Q}{t}=k_iA\frac{T_w-T_i}{l_i}$
$\frac{Ql_i}{tk_iA}=T_w-T_i$
$\frac{Q}{t}=k_bA\frac{T_o-T_w}{l_b}$
$\frac{Ql_b}{tk_bA}=T_o-T_w$
$\frac{Ql_i}{tk_iA}+\frac{Ql_b}{tk_bA}=T_w-T_i+T_o-T_w=35F^o$
$\frac{Q}{tA}(R_i+R_b)=35F^o$
$\frac{Q}{t}=\frac{(195ft^2)(35F^o)}{19ft^2hF^o/Btu+1ft^2hF^o/Btu}=340Btu/h$
$\frac{Q}{t}=\frac{(195ft^2)(35F^o)}{19ft^2hF^o/Btu+0.6ft^2hF^o/Btu}=350Btu/h$
