Asolar water heater for domestic hot-water supply uses solar collecting panels with a collection efficiency of 40.0 % in a location where the average solar-energy input is 200 w/m2.

a. if the water comes into the house at 16.0 ∘c and is to be heated to 60.0 ∘c, what volume of water can be heated per hour if the area of the collector is 29.5 m2?

v = l

46.19 L

Explanation:

The efficiency of the solar water heater is 40% which means 40% of the solar energy is converted to useful energy, ie. used to heat the water.

Useful energy = P = solar energy * available area * efficiency

P = 200 W/m^2 * 29.5 m^2 * 40%

P = 2360 W = 2.36 kJ/s

This means that 2.36 kJ of useful energy will be utilized per second. Converting this to the useful energy in hour gives us:

Average energy in one hour = 2.36 kJ/s * 3600 s/h = 8496 kJ

The specific heat capacity of water is 4.18 kJ/kg.C which means it will take 4.18 kJ of energy to raise the temperature of 1 kg of water by 1 degree C. Equating the energy change of the water for the given temperature rise and mass (unknown) to the useful energy utilized in one hour, we can solve to determine the unknown mass. This will give us the mass of water heated in one hour:

Energy = mass * specific heat capacity * (final temperature - initial temperature)

8496 = mass * 4.18 * (60 - 16)

mass = 46.19 kg

Lastly, this mass has to be converted to volume. Assuming density of water is constant through out the heating process:

volume = mass / density

volume = 46.19 kg / 1 kg/L

volume = 46.19 L

the answer would be a

wavelength divides in half and speed stays the same.