A house is designed to have passive solar energy features. Brickwork incorporated into the interior of the house acts as a heat absorber. Each brick weighs approximately \(1.8 \mathrm{~kg}\). The specific heat of the brick is \(0.85 \mathrm{~J} / \mathrm{g}-\mathrm{K} .\) How many bricks must be incorporated into the interior of the house to provide the same total heat capacity as \(1.7 \times 10^{3}\) gal of water?

Water has a density of 1 gram per milliliter (g/mL) or 1 kg per liter. We're given the volume in gallons, so first, we need to convert it to liters and then to mass. There are 3.78541 liters in a gallon, so: $$ 1.7\times10^3~\mathrm{gal} \times\frac{3.78541~\mathrm{L}}{1~\mathrm{gal}} = 6432.197~\mathrm{L} $$ As 1 L of water weighs 1 kg: $$ 6432.197~\mathrm{L} \times\frac{1~\mathrm{kg}}{1~\mathrm{L}} = 6432.197~\mathrm{kg} $$ So, the mass of water is 6432.197 kg.

Now, we'll calculate the total heat capacity of the water. We know the specific heat of water is \(4.18 \mathrm{~J/g} \mathrm{~K}\) or \(4180 \mathrm{~J/kg} \mathrm{~K}\). The heat capacity of water can be calculated using the formula: $$ Q_\mathrm{water} = m_\mathrm{water} \cdot c_\mathrm{water} $$ Where \(Q_\mathrm{water}\) is the heat capacity of water, \(m_\mathrm{water}\) is the mass of water, and \(c_\mathrm{water}\) is the specific heat of water. $$ Q_\mathrm{water} = 6432.197~\mathrm{kg} \times 4180~\mathrm{J/kg\cdot K} = 26,885,182.46~\mathrm{J/K} $$ So, the total heat capacity of the water is \(26,885,182.46~\mathrm{J/K}\).

We're given the specific heat of the brick, which is \(0.85 \mathrm{~J} / \mathrm{g}-\mathrm{K}\), and the mass of a single brick weighing 1.8 kg. We'll convert the specific heat to J/kg-K: $$ 0.85 \frac{\mathrm{J}}{\mathrm{g}\cdot\mathrm{K}} \times \frac{1000~\mathrm{g}}{1~\mathrm{kg}} = 850~\mathrm{J/kg\cdot K} $$ Now, we can calculate the heat capacity of a single brick using the formula: $$ Q_\mathrm{brick} = m_\mathrm{brick} \cdot c_\mathrm{brick} $$ Where \(Q_\mathrm{brick}\) is the heat capacity of one brick, \(m_\mathrm{brick}\) is the mass of one brick, and \(c_\mathrm{brick}\) is the specific heat of the brick. $$ Q_\mathrm{brick} = 1.8~\mathrm{kg} \times 850~\mathrm{J/kg\cdot K} = 1530~\mathrm{J/K} $$ So, the heat capacity of one brick is \(1530~\mathrm{J/K}\).

Now, we'll find the number of bricks needed to provide the same total heat capacity as the water. To do this, we'll simply divide the total heat capacity of the water by the heat capacity of one brick: $$ \mathrm{Number~of~bricks} = \frac{Q_\mathrm{water}}{Q_\mathrm{brick}} = \frac{26,885,182.46~\mathrm{J/K}}{1530~\mathrm{J/K}} = 17566.6 $$ As we can't have a fraction of a brick, we'll round up to the nearest whole number: $$ \mathrm{Number~of~bricks} = 17567 $$ So, 17,567 bricks must be incorporated into the interior of the house to provide the same total heat capacity as \(1.7 \times 10^{3}\) gal of water.