Posted by monique on .
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 450 L of water in the tank from 18°C to 50°C in 2.8 h when the intensity of incident sunlight is 520 W/m2? The specific heat of water is 4186 J/kg·K. The density of water is 1.00 g/cm3.

physics 
UTS student,
The intensity of sun light falling on the earth = I'
Let area of the plate = A
Power received by the plate, P = I'A
Efficiency of the over all system = 28 %
The power given to the tank is given by
P = 26% * I' A = 0.28 I'A...........(1)
m= mass of 450 L of water
S =specific heat of water = 4186 J/kg·K
4.186 J/g.K
ΔT = temparature difference = 32 deg C
The energy needed to raise the temp of water from 18 to 50 C is
Q = mSΔT = J
4.186*(450*10^3)*32
=60278400 J
power in put needed
P ' = Q / 1.0 * 60*60 sec
P =60278400 / 2.8*60*60 sec
P' = 5980 W (2)
From the equations (1) and (2) we have
0.28 I'A = 5980
A =5980/(0.28*520)
A = 41 m^2