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 11.0% (that is, 89% of the incident solar energy is lost from the system). What collector area (in square meters) is necessary to raise the temperature of 260 L of water in the tank from 25°C to 40°C in 1.0 h when the intensity of incident sunlight is 570 W/m2? The specific heat of water is 4186 J/kg·K. The density of water is 1.00 g/cm3.
Explanation / Answer
intensity of incident sunlight I = 570 W / m 2 specific heat of water c = 4186 J / kg . K change in temperature T = ( 40+273 ) - ( 25 + 273 ) = 15 K volume of water V = 260 L = 260 *10 ^ -3 m ^3 density of watre = 1000 kg / m 3 mass of water m = 0.260 *1000 = 260 kg energy released Q = m c T = 260 * 4186 * 15 = 1.6325*10^ 7 J power P = Q / time = 1.6325 *10 ^ 7 / 1 h = 1.6325*10 ^ 7 / 3600 = 4.53483 *10^ 3 W the efficiency of overal system is 11 % 0.11 I = P / A Area A = 4.53483 *10 ^ 3 / 0.11* 570 = 72 m ^ 2