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 24.0% (that is, 76% of the incident solar energy is lost from the system). What collector area (in square meters) is necessary to raise the temperature of 330 L of water in the tank from 21°C to 46°C in 2.7 h when the intensity of incident sunlight is 540 W/m2? The specific heat of water is 4186 J/kg·K. The density of water is 1.00 g/cm3.
To find the collector area necessary, we can use the equation:
Energy absorbed by water = Energy incident on the collector - Energy lost by the system
First, let's calculate the energy absorbed by the water, considering the specific heat of water and the temperature change:
Energy absorbed by water = (mass of water) x (specific heat of water) x (change in temperature)
Given that the density of water is 1.00 g/cm^3 and the volume of water is 330 L, we need to convert the volume to mass:
Mass of water = (volume of water) x (density of water)
Mass of water = (330 L) x (1000 cm^3/L) x (1.00 g/cm^3)
Mass of water = 330,000 g = 330 kg
The change in temperature is 46°C - 21°C = 25°C.
Energy absorbed by water = (330 kg) x (4186 J/kg·K) x (25°C)
Next, let's calculate the energy incident on the collector:
Energy incident on the collector = (intensity of sunlight) x (area of the collector) x (time)
Given that the intensity of sunlight is 540 W/m^2 and the time is 2.7 h, we need to convert the time to seconds:
Time = 2.7 h x 3600 s/h = 9720 s
Energy incident on the collector = (540 W/m^2) x (area of the collector) x (9720 s)
Now, we can calculate the energy lost by the system:
Energy lost by the system = (1 - efficiency) x (energy incident on the collector)
Given that the efficiency is 24%, we need to convert it to decimal form:
Efficiency = 24% = 0.24
Energy lost by the system = (1 - 0.24) x (energy incident on the collector)
Now, we can equate the energy absorbed by the water to the energy incident on the collector minus the energy lost by the system:
(330 kg) x (4186 J/kg·K) x (25°C) = (540 W/m^2) x (area of the collector) x (9720 s) - (1 - 0.24) x (energy incident on the collector)
Simplifying the equation:
(330 kg) x (4186 J/kg·K) x (25°C) = (540 W/m^2) x (area of the collector) x (9720 s) - (0.76) x (540 W/m^2) x (area of the collector) x (9720 s)
Now, we can solve for the area of the collector:
(330 kg) x (4186 J/kg·K) x (25°C) = (540 W/m^2) x (area of the collector) x (9720 s) x (1 - 0.76)
Simplifying further:
(330 kg) x (4186 J/kg·K) x (25°C) = (540 W/m^2) x (area of the collector) x (9720 s) x (0.24)
To find the area of the collector, we can rearrange the equation:
Area of the collector = [(330 kg) x (4186 J/kg·K) x (25°C)] / [(540 W/m^2) x (9720 s) x (0.24)]
Calculating the expression:
Area of the collector = [(330 kg) x (4186 J/kg·K) x (25°C)] / [(540 W/m^2) x (9720 s) x (0.24)]
Now, you can input the values into the equation to find the area of the collector in square meters.