Question

A passive solar house that is losing heat to the outdoors at an average rate of \(50,000 \mathrm{kJ} / \mathrm{h}\) is maintained at \(22^{\circ} \mathrm{C}\) at all times during a winter night for \(10 \mathrm{h}\). The house is to be heated by 50 glass containers each containing \(20 \mathrm{L}\) of water that is heated to \(80^{\circ} \mathrm{C}\) during the day by absorbing solar energy. A thermostat-controlled \(15-\mathrm{kW}\) back-up electric resistance heater turns on whenever necessary to keep the house at \(22^{\circ} \mathrm{C}\). \((a)\) How long did the electric heating system run that night? (b) How long would the electric heater run that night if the house incorporated no solar heating? \(\quad\)

Short answer

Answer: a) To find the time the electric heating system runs when passive solar heating is used, first calculate the energy provided by the solar heating system using the given data. Then, calculate the heat loss during a 10-hour winter night. Next, calculate the energy needed by the electric heating system by finding the difference between the heat loss and the energy provided by solar heating. Finally, calculate the running time by dividing the energy needed by the power of the electric heating system (15 kW). b) To find the time the electric heating system runs when passive solar heating is not used, directly calculate the heat loss during a 10-hour winter night. Then, calculate the running time by dividing the heat loss by the power of the electric heating system (15 kW).

Step-by-step solution

Calculate the energy provided by the solar heating system.

To find this, we need to find the energy provided by the 50 glass containers of water, each with 20 L of water heated to 80°C during the day. We know that the specific heat capacity of water is \(4186 \frac{\mathrm{J}}{\mathrm{kg \cdot K}}\). First, we convert 20 liters of water to kilograms assuming that the density of water is 1 \(\frac{\mathrm{kg}}{\mathrm{L}}\): \(20 \mathrm{L} \times 1 \frac{\mathrm{kg}}{\mathrm{L}} = 20 \mathrm{kg}\) For each container: \(Q = mc \Delta T\) where: \(Q\) = heat energy \(m\) = mass of water \(c\) = specific heat capacity of water \(\Delta T\) = change in temperature \(\Delta T = 80^{\circ} \mathrm{C} - 22^{\circ} \mathrm{C} = 58^{\circ} \mathrm{C}\) Now, we calculate the heat energy in a single container: \(Q = 20 \mathrm{kg} \times 4186 \frac{\mathrm{J}}{\mathrm{kg \cdot K}} \times 58 \mathrm{K}\) Then, the total energy provided by all 50 containers: \(Q_{total} = 50 \times Q\)

Calculate heat loss during a winter night.

We are given that the house loses heat at a rate of \(50000 \mathrm{kJ/h}\), and the heater runs for 10 hours. To find the total heat loss in Joules, we can use the following formula: \(Q_{loss} = \textrm{heat loss rate} \times \textrm{duration}\) Convert the heat loss rate to Joules: \(50000 \mathrm{kJ/h} \times \frac{1 \mathrm{h}}{3600 \mathrm{s}} = \frac{50000}{3.6} \mathrm{kJ/s} = 13888.89 \mathrm{J/s}\) Calculate the total heat loss during a winter night: \(Q_{loss} = 13888.89 \mathrm{J/s} \times 10 \mathrm{h} \times 3600 \frac{\mathrm{s}}{\mathrm{h}}\)

Calculate how long the electric heating system run when passive solar heating is used.

The energy provided by the passive solar heating system should be used up first, and then the electric heating system kicks in. We need to find the difference between the heat loss and the energy provided by solar heating: \(Q_{needed} = Q_{loss} - Q_{total}\) The 15-kW electric heating system will provide the required energy, so we find the running time as follows: \(t_{used} = \frac{Q_{needed}}{15000 \mathrm{W}}\)

Calculate how long the electric heating system run when passive solar heating is not used.

In this case, all the heat loss should be compensated by the electric heating system. The running time can be found using the same formula as in Step 3: \(t_{no\_solar} = \frac{Q_{loss}}{15000 \mathrm{W}}\) Now, we have found how long the electric heating system runs during a winter night when passive solar heating is used and when it is not used.

Key concepts

Thermal Energy Transfer

Thermal energy transfer is a fundamental concept in understanding how passive solar heating works. The basic principle revolves around the movement of heat from a warmer area to a cooler one until there is no longer a temperature difference. This transfer can occur in three ways: conduction, convection, and radiation. In the context of a passive solar house, the sun's radiation provides thermal energy that gets absorbed and stored in materials within the home, such as water in glass containers. During cooler times, like a winter night, this stored heat is gradually released into the home, offsetting the heat lost to the outside environment.

Effective thermal energy transfer in passive solar design can significantly reduce the need for artificial heating, as seen in the exercise where water stores heat energy during the day. This heat is then naturally released to maintain a comfortable temperature inside when it's colder outside, showcasing direct utilization of energy transfer.

Specific Heat Capacity

Specific heat capacity is a property that quantifies the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin). The specific heat capacity of water is relatively high at approximately \(4186 \frac{\mathrm{J}}{\mathrm{kg \cdot K}}\), meaning water can store a significant amount of heat energy without experiencing a large temperature change.

In our exercise, the ability of water to store thermal energy during the day and release it at night is a perfect example of tapping into this physical property. By understanding and utilizing the specific heat capacity of water, the design of the passive solar heating system can be optimized. The containers of water in the exercise are a type of 'thermal battery', capturing heat energy when it's available and releasing it to maintain temperature, thus reducing the reliance on electric heaters.

Electric Resistance Heater Efficiency

The efficiency of an electric resistance heater is a measure of how well it converts electrical energy into thermal energy. In an ideal scenario, these heaters are considered to be 100% efficient since nearly all the electrical energy is converted into heat. However, the overall heating efficiency of a building isn't just about the heater's efficiency; it's also about how well the building retains the heat produced and minimizes heat loss.

In the exercise, the thermostat-controlled electric resistance heater supplements the passive solar heating. Its efficiency comes into play when ensuring that the temperature is maintained at the desired level without unnecessary energy waste. When solar heating is insufficient, the electric heater compensates for the heat loss. The usage time of this heater reflects the overall thermal performance of the house; the less it runs, the better the house is at retaining heat, resulting in both energy and cost savings.