In a solar water heater, energy from the Sun is gathered by water that circulates through tubes in a rooftop collector. The solar radiation enters the collector through a transparent cover and warms the water in the tubes; this water is pumped into a holding tank. Assume that the efficiency of the overall system is 15.0% (that is, 85% of the incident solar energy is lost from the system). What collector area (in square meters) is necessary to raise the temperature of 440 L of water in the tank from 24°C to 36°C in 2.6 h when the intensity of incident sunlight is 610 W/m2? The specific heat of water is 4186 J/kg·K. The density of water is 1.00 g/cm3
To calculate the collector area necessary to raise the temperature of the water in the tank, we need to calculate the amount of energy required to heat the water and then determine the amount of solar energy incident on the collector.
First, let's determine the energy required to heat the water. The energy equation for heating a substance is given by:
Energy = mass × specific heat × temperature change
In this case, we have:
Mass of water = 440 L = 440 kg (since the density of water is 1.00 g/cm3)
Specific heat of water = 4186 J/kg·K
Temperature change = 36°C - 24°C = 12°C
Using the equation, we can compute the energy required:
Energy = 440 kg × 4186 J/kg·K × 12°C
Energy = 22,047,360 J
Next, let's determine the amount of solar energy incident on the collector. We have the intensity of incident sunlight, which is given as 610 W/m2. The time of exposure is 2.6 hours.
To calculate the energy incident on the collector, we multiply the intensity of sunlight by the collector area and the time of exposure:
Energy incident = intensity of sunlight × collector area × time of exposure
We know that the system efficiency is 15.0% or 0.15. Therefore, the energy incident on the collector is only 0.15 of the total solar energy incident on the system:
Energy incident on collector = efficiency × energy incident
We can rearrange the equation to solve for the required collector area:
Collector area = (energy required) / (efficiency × intensity of sunlight × time of exposure)
Substituting the values we calculated earlier:
Collector area = 22,047,360 J / (0.15 × 610 W/m2 × 2.6 hours)
Collector area = 22,047,360 J / (0.15 × 610 W/m2 × 2.6 hours) = 235.42 m2
Therefore, the collector area necessary to raise the temperature of 440 L of water from 24°C to 36°C in 2.6 hours is approximately 235.42 square meters.