Problem 38 - College Physics

Question

Solar water heater. A solar water heater for domestic hot- water supply uses solar collecting panels with a collection efficiency of 50\(\%\) in a location where the average solar-energy input is 200 \(\mathrm{W} / \mathrm{m}^{2} .\) If the water comes into the house at \(15.0^{\circ} \mathrm{C}\) and is to be heated to \(60.0^{\circ} \mathrm{C},\) what volume of water can be heated per hour if the area of the collector is 30.0 \(\mathrm{m}^{2} ?\)

Step-by-Step Solution

Answer

Approximately 57.34 liters of water can be heated per hour.

Step 1: Calculate the Effective Power Collected

First, determine the effective power collected by the solar panels. The panels have an efficiency of 50%, and the solar energy input is 200 W/m². The area of the collector is 30 m². Compute the raw power received by multiplying the solar energy input by the area, then apply the efficiency: \[ P_{eff} = 0.5 \times 200 \, \text{W/m}^2 \times 30 \, \text{m}^2 = 3000 \, \text{W} \].

Step 2: Determine the Energy Needed to Heat Water

Calculate the amount of energy required to raise the temperature of the water from 15°C to 60°C. Use the specific heat capacity of water, which is approximately 4186 J/kg°C. Assume 1 liter of water has a mass of 1 kg. The energy required for each kilogram is given by: \[ Q = m \times c \times \Delta T = 1 \, \text{kg} \times 4186 \, \text{J/kg°C} \times (60°C - 15°C) = 188370 \, \text{J} \].

Step 3: Calculate the Mass of Water Heated per Hour

Determine how much energy the collector can gather in one hour, then use it to find out how much water can be heated. Since power is energy per unit time, the energy collected per hour is: \[ E_{hour} = P_{eff} \times 3600 \, \text{s} = 3000 \, \text{W} \times 3600 \, \text{s} = 10,800,000 \, \text{J} \].

Step 4: Determine the Volume of Water Heated

Finally, calculate the volume of water that can be heated using the energy collected. Since each kilogram of water requires 188370 J, the mass of water that can be heated is: \[ m = \frac{10,800,000 \, \text{J}}{188370 \, \text{J/kg}} \approx 57.34 \, \text{kg} \]. Hence, the volume of water is approximately 57.34 liters since 1 kg of water is roughly equivalent to 1 liter.

Key Concepts

Solar Energy Efficiency

Solar energy efficiency refers to the effectiveness of solar panels in converting sunlight into usable energy. It is crucial to consider because it impacts how much energy we can harness. For a solar water heater, the efficiency indicates the percentage of solar energy converted into heat. This is often expressed as a percentage.

In the provided problem, the solar collectors have an efficiency of 50%. This means that out of all the sunlight that strikes the panels, only half is converted into usable energy. The rest is lost due to various factors such as reflection or heat dissipation. Understanding this concept helps in calculating the real-world power output from a solar panel. Knowing this efficiency allows us to accurately calculate the effective power, which is a pivotal step in energy conversion calculations.

It's also important to note that efficiency can be affected by various factors, such as the angle of the panels, the weather conditions, and even the cleanliness of the panel's surface. Hence, improving these conditions might lead to better efficiency and energy output.

Specific Heat Capacity of Water

The specific heat capacity of a substance is the amount of heat required to change the temperature of one kilogram of the substance by one degree Celsius. For water, this value is quite high at approximately 4186 J/kg°C. This means it takes 4186 joules to raise 1 kilogram of water by 1°C.

Why is this important for a solar water heater? Because it determines how much energy is needed to heat the water to the desired temperature. In our problem, the water's temperature needs to rise from 15°C to 60°C. Understanding the specific heat capacity allows us to calculate the exact energy required for this temperature change.

Given the specific heat capacity, we can calculate the total energy needed to heat a certain volume of water, allowing us to plan and optimize the energy use effectively. This concept is fundamental in designing solar water heating systems that efficiently use available solar energy to achieve desired heating goals.

Energy Conversion Calculations

Energy conversion calculations are essential in determining how energy sourced from solar panels is utilized for heating water. In the problem, this involves calculating several aspects:

• First, compute the effective power collected, using the formula: \( P_{eff} = \text{Efficiency} \times \text{Solar energy input} \times \text{Area} \). For this scenario, the panels gathered an effective power of 3000 W.
• Then, calculate the energy needed to heat water using the formula: \( Q = m \times c \times \Delta T \). Given in the exercise, the energy required per kilogram of water is 188370 J.
• Finally, find the mass of the water that can be heated using the energy collected. This can be done with the formula: \( m = \frac{E_{hour}}{Q} \), where \( E_{hour} \) is the energy collected per hour, calculated as 10,800,000 J.

By understanding these calculations, students can better comprehend how solar energy and water heating processes intersect. They showcase how energy from the sun is converted and utilized efficiently to achieve practical goals like heating water for domestic use.