Question

One method of passive solar heating is to stack gallons of liquid water inside the buildings and expose them to the sun. The solar energy stored in the water during the day is released at night to the room air, providing some heating. Consider a house that is maintained at \(22^{\circ} \mathrm{C}\) and whose heating is assisted by a 350 -L water storage system. If the water is heated to \(45^{\circ} \mathrm{C}\) during the day, determine the amount of heating this water will provide to the house at night. Assuming an outside temperature of \(5^{\circ} \mathrm{C},\) determine the exergy destruction associated with this process.

Short answer

Answer: The heat transfer provided by the water storage system is 33,532 kJ, and the exergy destruction associated with this process is 96.203 kJ/K.

Step-by-step solution

Calculate the heat transfer

To find the amount of heat transfer, we will use the specific heat equation: $$Q = m \times C \times \Delta T$$ where \(Q\) is the heat transfer, \(m\) is the mass of water, \(C\) is the specific heat of water, and \(\Delta T\) is the temperature difference. The volume of water given is 350 L, and the density of water is 1000 kg/m³. So, $$m = \text{density} \times \text{volume} = 1000 \times 0.35 = 350\,\text{kg}$$ The specific heat of water is \(C = 4.186\,\text{kJ/kg}^{\circ}\text{C}\). The temperature difference is \(\Delta T = 45^{\circ}\text{C} - 22^{\circ}\text{C} = 23^{\circ}\text{C}\). Now, we can calculate the heat transfer: $$Q = 350\,\text{kg} \times 4.186\,\text{kJ/kg}^{\circ}\text{C} \times 23^{\circ}\text{C} = 33532\,\text{kJ}$$

Calculate the exergy destruction

To find the exergy destruction, we will use the formula for exergy destruction: $$\Delta E_{\text{dest}} = \frac{Q}{T_{\text{hot}}} - \frac{Q}{T_{\text{cold}}}$$ where \(\Delta E_{\text{dest}}\) is the exergy destruction, \(Q\) is the heat transfer, \(T_{\text{hot}}\) is the temperature of the hot water, and \(T_{\text{cold}}\) is the temperature of the outside surroundings. Convert the temperatures to Kelvin: $$T_{\text{hot}} = 45^{\circ}\text{C} + 273.15 = 318.15\,\text{K}$$ $$T_{\text{cold}} = 5^{\circ}\text{C} + 273.15 = 278.15\,\text{K}$$ Now, we can calculate the exergy destruction: $$\Delta E_{\text{dest}} = \frac{33532\,\text{kJ}}{318.15\,\text{K}} - \frac{33532\,\text{kJ}}{278.15\,\text{K}} = 24.243\,\text{kJ/K} - 120.446\,\text{kJ/K} = -96.203\,\text{kJ/K}$$ However, exergy destruction is always a positive value, so we take its absolute value: $$\Delta E_{\text{dest}} = |-96.203\,\text{kJ/K}| = 96.203\,\text{kJ/K}$$ So, the amount of heating this water will provide to the house at night is 33532 kJ, and the exergy destruction associated with this process is 96.203 kJ/K.

Key concepts

Heat Transfer Calculation

When looking at solar heating thermodynamics, one essential concept is the heat transfer calculation. In practice, you might find a scenario where water stored in a solar heating setup is used to regulate the temperature inside a building. To quantify the heat provided by this system, a straightforward yet powerful equation is utilized:
\[ Q = m \times C \times \Delta T \] Where Q stands for the amount of heat transfer, m is the mass of water, C is the specific heat of water, and \Delta T represents the change in temperature. Considering a case where the mass of water is 350 kg (from 350 L, assuming 1 kg/L density), and the temperature change is from 22°C to 45°C, applying the specific heat capacity of water yields the total heat energy released.

This kind of calculation applies well beyond the classroom; it's a critical step for engineers and environmental scientists designing energy-efficient homes or assessing renewable energy sources.

Exergy Destruction

Another essential component of solar heating thermodynamics is exergy destruction. Exergy is a measure of a system's ability to do work when it is brought into equilibrium with its surroundings. In the context of solar heating, exergy destruction refers to the irreversible loss of potential energy due to temperature differences.

To calculate exergy destruction when water releases heat, you might use the formula: \br \[ \Delta E_{\text{dest}} = \frac{Q}{T_{\text{hot}}} - \frac{Q}{T_{\text{cold}}} \] where Q is the heat transferred, T_{\text{hot}} and T_{\text{cold}} refer to the temperatures of the hot water and the outside environment, respectively. This concept underlines that while energy can never be destroyed according to the law of conservation of energy, exergy, which is the usable energy, can, highlighting the inherent inefficiencies in thermal systems.

Specific Heat Capacity

A key term in understanding solar heating thermodynamics is specific heat capacity, often simply called specific heat. It is the amount of heat one gram of a substance must absorb or lose to change its temperature by one degree Celsius. For water, the specific heat capacity is typically around 4.186 joules per gram per degree Celsius (kJ/kg°C).

When calculating the heat transfer for a solar heating system, knowledge of specific heat is crucial. It ensures accurate predictions of the system’s thermal storage capability. In the given exercise, understanding that the specific heat of water is relatively high compared to other substances allows us to appreciate the effectiveness of water as a storage medium for solar energy. This property enables water to store large amounts of heat without significant changes in temperature, making it ideal for passive solar heating applications.