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 is necessary to raise the temperature of 230 l of water in the tank from 20°c to 49°c in 2.6 h when the intensity of incident sunlight is 500 w/m2? the specific heat of water is 4186 j/kg·k. the density of water is 1.00 g/cm3.

21.30 m2

Explanation:

First of all, we calculate the energy necessary to increase the temperature of the water from 20°C to 49°C. We use the specific heat in order to achieve this:

Where m_w is the mass of water, sh is the specific heat.

The mass of the water can be found using the volume of water and density:

Then:

You want to heat the water in 2.6h, this means, you will need to supply a power equal to the total amount of energy required divided by the time in seconds. In 2.6h there are 2.6*3600 = 9360s:

But remember, you have an efficiency of 28%, so the actual amount of power that you need to recieve from the sun is equal to:

Now, for the power of a collector is:

Where Psun is the power of the sun, S is the intensity of incident sunlight and A is the area. Solving for Area:

5 w for 1hr = 5x60x60 joules=300x60=3000x6=18,000j as total energy conversion. efficiency of bulb needed for radiant to thermal within 18000

answer; equivalent to a one(1) horsepower motor;