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 28.0% (that is, 72% of the incident solar energy is lost from the system). What collector area (in square meters) is necessary to raise the temperature of 340 L of water in the tank from 19°C to 44°C in 1.8 h when the intensity of incident sunlight is 630 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 = 630 W / m 2 specific heat of water c = 4186 J / kg . K change in temperature T = ( 44+273 ) - ( 19 + 273 ) = 25 K volume of water V = 340 L = 340 *10 ^ -3 m ^3 density of watre = 1000 kg / m 3 mass of water m = 0.340 *1000 = 340 kg energy released Q = m c T = 340 * 4186 * 25 = 3.5581*10^ 7 J power P = Q / time = 3.5581 *10 ^ 7 / 1 h = 3.5581*10 ^ 7 / 3600 =9.883 *10^ 3 W the efficiency of overal system is 28 % 0.11 I = P / A Area A = 9.883 *10 ^ 3 / 0.11* 630 = 142.611 m ^ 2