Problem 105 - Chemistry: The Central Science in SI Units, Global Edition

Question

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?

Step-by-Step Solution

Answer

To provide the same total heat capacity as \(1.7 \times 10^{3}\) gal of water, 17,567 bricks must be incorporated into the interior of the house.

Step 1: Convert the volume of water to mass

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.

Step 2: Calculate the total heat capacity of water

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 \(26,885,182.46~\mathrm{J/K}\).

Step 3: Calculate the heat capacity of one brick

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 \(1530~\mathrm{J/K}\).

Step 4: Calculate the number of bricks required

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.

Key Concepts

Specific Heat Capacity

Specific heat capacity is a property of a material that tells us how much heat energy is needed to raise the temperature of a certain mass of the substance by 1-degree Celsius (or Kelvin).

It's a crucial concept in understanding how different materials absorb and store heat. For brick, the specific heat capacity is given as \(0.85 \mathrm{~J/g\cdot K}\), which means that for each gram of brick, 0.85 Joules are needed to increase the temperature by 1 K.

To make calculations easier, we often convert it to \(\mathrm{J/kg\cdot K}\):

• Multiply by 1000 to switch from grams to kilograms.

This makes it align with standard units used for other substances, like water, whose specific heat capacity is \(4180 \mathrm{~J/kg\cdot K}\). Understanding this helps us compare how efficiently different materials can store and transfer heat.

Heat Absorber

A heat absorber is a material that actively takes in thermal energy from its surroundings. In passive solar systems, materials like brick are used because of their ability to absorb and gradually release heat.

The brick in our example absorbs sunlight during the day. As the temperature drops at night, it slowly releases this absorbed heat, warming the interior space. This process reduces the need for other heating methods.

Key points about heat absorbers include:

• They can help stabilize indoor temperatures.
• They are cost-effective and environmentally friendly.

Using materials like brickwork is a sustainable choice in building designs.

Thermal Mass

Thermal mass refers to a material's ability to absorb, store, and release heat energy. It's an essential component in passive solar heating systems.

When a material has high thermal mass, it captures heat efficiently during the day and releases it slowly at night, smoothing out temperature fluctuations.

Ideal characteristics of thermal mass include:

• High density and specific heat capacity.
• Ability to remain effective over long periods.

In the example, bricks act as thermal mass. They moderate temperature changes by storing heat energy during the day and releasing it when it becomes cooler. This makes living spaces more comfortable without using additional energy.

Heat Capacity Calculation

Heat capacity is the total amount of heat energy required to change a substance's temperature by a certain amount. Calculating heat capacity helps us understand and design systems like passive solar homes.

The formula for heat capacity \(Q\) is based on the mass \(m\) and specific heat capacity \(c\) of the material:
\[ Q = m \cdot c \]

In our example, we calculate the heat capacity of water and bricks:

• First, find the mass, using density for water or direct measurements for bricks.
• Then, use the formula to find the heat capacity.

Knowing the heat capacity allows us to determine how many bricks are needed to match the thermal storage of water.

It provides a practical way to balance heat storage in building materials effectively.