This refers to passive solar collector, where solar radiation is used to heat water. It is therefore not on the solar cells.

What characteristics should solar collectors have, to function in an optimal way?
Why should these properties be fulfilled?

If there is a water flow of 500 g / minute to a solar panel on one m2, how much water is heated up under optimal conditions during the passage of the collector? Expect that the collector is in Gothenburg and the calculation is done for one day in mid-June (noon)

It makes no sense to use the noon position of the sun to predict the full-day solar energy received by a collector. The middle of June is very close to summer solstice time in the northern hemisphere. The latitude of Gothenburg is 58 degrees N. At noon on the solstice, the sun will be 90-58+23 = 55 degrees above the horizon. For optimum collection at that time, the solar panels should be tilted to the south by about 35 degrees. Then the sun's rays will be perpendicular to the panels at noon.

That will not be the optimum panel tilt angle for all times of day and days of the year, however.

There is a good discussion of this subject at
http://www.macslab.com/optsolar.html

For a single tilt that will not be changed during the year, the best angle is the latitude, times 0.75, plus 3 degrees. That would be about 46 degrees in Gothenburg

how should i consider the flow of water

500 g / minute and the area of the collector

500 g/minute is a very high flow rate for a home solar water heater. How much do you need to heat the water in a single pass?

For a single pass through the solar collector,

(mass flow rate)*(specific heat of water)*(delta T) = (Solar heating per area)*(collector area) - convection losses.

Convetion losses become high when you try to heat the water more than 10 degrees C above the ambient temperature.

The solar heating per area will be about 1000 W/m^2 times a cosine factor, if the sun is out.

To function optimally, solar collectors should have the following characteristics:

1. Absorptivity: Solar collectors should have high absorptivity, meaning they should effectively absorb solar radiation. This is usually achieved by using materials with dark colors or coatings that can absorb a large percentage of the incoming solar energy.

2. Low emissivity: The collectors should have low emissivity, meaning they should minimize the amount of heat radiation they release back into the environment. This is usually achieved by using materials or coatings that have low thermal emissivity.

3. Thermal conductivity: The collectors should have high thermal conductivity, meaning they should efficiently transfer the absorbed heat to the fluid flowing through them. This is usually achieved by using materials with high thermal conductivity, such as metals or special heat transfer fluids.

4. Transparency: If the solar collector has a cover or glazing, it should be transparent to allow the solar radiation to pass through without significant losses. Some collectors use transparent materials, such as glass or polymers, to maximize the solar energy reaching the absorber.

These properties should be fulfilled because they directly impact the efficiency and effectiveness of the solar collector system. By optimizing these characteristics, more solar radiation can be converted into usable heat energy, resulting in higher overall system efficiency.

Now, let's calculate the amount of water heated up under optimal conditions for the given scenario:

Assuming the water flow rate is 500 g/minute, we need to convert it to kg/second for calculations.
500 g/minute = 0.5 kg/minute = (0.5/60) kg/second ≈ 0.0083 kg/second.

To calculate the amount of water heated up, we need to consider the solar irradiance in Gothenburg at noon in mid-June. According to solar insolation data, we can assume an average value of 600 watts per square meter (W/m^2) for this time and location.

Let's assume the collector area is 1 m^2.

The amount of heat transferred to the water can be calculated using the formula:

Q = m * c * ΔT

where:
Q = heat transferred (in joules)
m = mass of water (in kg)
c = specific heat capacity of water (4.186 J/g°C)
ΔT = temperature increase (in °C)

Considering the water flow rate, the mass of water heated per second can be calculated as:

m = water flow rate (kg/second) * time (seconds)
m = 0.0083 kg/second * 24 hours * 60 minutes * 60 seconds
m ≈ 7.15 kg

Assuming an average temperature increase of ΔT = 10 °C:

Q = 7.15 kg * 4.186 J/g°C * 10 °C
Q ≈ 298.7 kJ

Therefore, under optimal conditions, approximately 298.7 kilojoules of heat energy would be transferred to the water passing through the collector in one day at noon in mid-June in Gothenburg.