Determine the winter \(R\)-value and the \(U\)-factor of a masonry wall that consists of the following layers: \(100-\mathrm{mm}\) face bricks, 100 -mm common bricks, \(25-\mathrm{mm}\) urethane rigid foam insulation, and \(13-\mathrm{mm}\) gypsum wallboard.

To find the \(R\)-values for each layer of the wall, we can use the following formula: $$R = \frac{d}{k}$$ Where \(R\) is the thermal resistance, \(d\) is the thickness of the material in meters, and \(k\) is the material's conductivity in \(W/(m\cdot K)\). We will need to find the appropriate \(k\) values for each material on our wall using a reference or source for material conductivity values. The thicknesses are given in the problem statement, but we will need to convert them from millimeters to meters. $$ d_{face\, bricks} =0.1\,\mathrm{m} \\ d_{common\, bricks} =0.1\,\mathrm{m} \\ d_{urethane\, rigid\, foam} =0.025\,\mathrm{m} \\ d_{gypsum\, wallboard} =0.013\,\mathrm{m} $$ To find the \(R\)-value for each layer, we can use the formula given above: $$R_{face\, bricks} = \frac{d_{face\, bricks}}{k_{face\, bricks}}$$ $$R_{common\, bricks} = \frac{d_{common\, bricks}}{k_{common\, bricks}}$$ $$R_{urethane\, rigid\, foam} = \frac{d_{urethane\, rigid\, foam}}{k_{urethane\, rigid\, foam}}$$ $$R_{gypsum\, wallboard} = \frac{d_{gypsum\, wallboard}}{k_{gypsum\, wallboard}}$$

To obtain the total thermal resistance of the wall, we can simply add the values we found for each layer: $$R_{total} = R_{face\, bricks} + R_{common\, bricks} + R_{urethane\, rigid\, foam} + R_{gypsum\, wallboard}$$

The \(U\)-factor, or thermal transmittance, is the inverse of the total \(R\)-value. So, to calculate the \(U\)-factor for the masonry wall, simply take the reciprocal of the total \(R\)-value: $$U = \frac{1}{R_{total}}$$ Now you have determined both the winter \(R\)-value and the \(U\)-factor for the masonry wall. Keep in mind that these values are approximate and may vary depending on factors like material quality, temperature variations and other external conditions.