Question

A passive solar house that is losing heat to the outdoors at \(3^{\circ} \mathrm{C}\) 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 \(15 \mathrm{kW}\) backup electric resistance heater turns on whenever necessary to keep the house at \(22^{\circ} \mathrm{C}\). Determine how long the electric heating system was on that night and the amount of entropy generated during the night.

Short answer

Answer: The electric heating system was on for approximately 4 hours and 57 minutes (17,807.04 seconds) during the night, and the amount of entropy generated was 2,514.98 kJ/K.

Step-by-step solution

Calculate the energy lost by the house during the night

The house loses heat at a rate of 50,000 kJ/h and the night lasts 10 hours. The energy lost by the house during the night can be calculated as follows: Energy lost = (Heat loss rate per hour) × (Duration of the night) Energy lost = 50,000 kJ/h × 10 h = 500,000 kJ

Calculate the energy provided by the water in the glass containers

We have 50 glass containers, each containing 20 L of water heated to 80°C. First, let's find out the mass of the water in all containers. The density of water is 1 kg/L, so the mass can be found as follows: Total mass of water = (Volume of water per container) × (Number of containers) × (Density) Total mass of water = 20 L × 50 × 1 kg/L = 1000 kg Next, let's find the temperature difference between the initial and final temperatures of the water: ΔT = (Final temperature) - (Initial temperature) ΔT = 22°C - 80°C = -58°C Now we can calculate the energy provided by the water in the glass containers using the specific heat capacity of water (c_water = 4.18 kJ/kg°C): Energy provided = (Total mass of water) × (Specific heat capacity) × (Temperature difference) Energy provided = 1000 kg × 4.18 kJ/kg°C × (-58°C) = -241,960 kJ

Calculate the energy required from the electric heater

The electric heater needs to make up for the difference between the energy lost by the house and the energy provided by the water. The energy required from the electric heater can be found as follows: Energy required = (Energy lost by the house) - (Energy provided by water) Energy required = 500,000 kJ - (-241,960 kJ) = 741,960 kJ

Determine how long the electric heating system was on

The backup electric resistance heater has a power of 15 kW. We can find the time the heater was on by dividing the energy required by the power of the heater: Time on = (Energy required) / (Power of the heater) Time on = 741,960 kJ / 15,000 W = 741.96 * (3600s / 15,000 W) = 17807.04 s So, the electric heating system was on for 17,807.04 seconds or approximately 4 hours and 57 minutes.

Calculate the amount of entropy generated

The heater's energy is transformed into heat at a constant temperature of 22°C. The heat transfer to the house increases the entropy. The amount of entropy generated can be calculated as follows: Entropy change = (Energy transferred as heat) / (Temperature in Kelvin) Entropy change = 741,960 kJ / (22°C + 273)K Entropy change = 741,960 kJ / 295 K = 2,514.98 kJ/K The amount of entropy generated during the night is 2,514.98 kJ/K.