Answer:
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
