The luminosity of the Sun is 4*10^33 erg/s, and its radius is 7*10^10 cm.
You are tasked with building a solar power plant in the Arizona desert, using solar panels with 10% efficiency. How large an area (km^2) must your solar panels cover to match the power output of a large nuclear powerplant (about a GigaWatt)? Please enter your answer in units of km^2
Well, well, building a solar power plant in the Arizona desert, are we? That's a bright idea! Let's crunch some numbers to figure out the area you need.
First, let's convert that GigaWatt into erg/s to keep things consistent. 1 GigaWatt is equal to 10^9 Watts, and 1 Watt is equal to 10^7 erg/s. So a GigaWatt is about 10^9 x 10^7 = 10^16 erg/s.
Now, the efficiency of your solar panels is 10%, which means only 10% of the Sun's luminosity will actually be converted into usable energy. So you need to multiply the solar panel efficiency by the luminosity of the Sun.
10% of 4*10^33 erg/s is 4*10^32 erg/s.
To calculate the area needed, we'll divide the power output of the solar panels by the power output of a single solar panel (taking efficiency into account).
So, the area required in square centimeters would be:
Area = (4*10^32 erg/s) / (10% of 4*10^33 erg/s)
That simplifies to:
Area = 4*10^32 erg/s / (0.1 x 4*10^33 erg/s)
Canceling out the common terms, we get:
Area = 1 / 0.1 = 10
Therefore, you need an area of 10 square centimeters. But since you prefer your answer in units of km^2, let's convert that.
If we consider 1 square centimeter as 10^4 square meters, then 10 square centimeters would be 10^5 square meters, which is equal to 0.01 km^2.
So, drumroll, please! You would need your solar panels to cover an area of 0.01 km^2 to match the power output of a large nuclear power plant.
Now, let's go shine a light on renewable energy!
To find the area that the solar panels must cover to match the power output of a large nuclear power plant, we need to calculate the power output of the solar panels.
Given:
Luminosity of the Sun = 4*10^33 erg/s
Radius of the Sun = 7*10^10 cm
Efficiency of solar panels = 10%
First, we need to calculate the total power output of the solar panels:
Power output of the solar panels = (Efficiency of solar panels) x (Luminosity of the Sun)
Power output of the solar panels = (10%)(4*10^33 erg/s)
Now, we need to convert the power output from erg/s to Watts, as the power output of the nuclear power plant is given in Watts.
1 Watt = 10^7 erg/s
So, the power output of the solar panels in Watts can be calculated as:
Power output of the solar panels = (10%)(4*10^33 erg/s) / (10^7 erg/s)
Now, we can compare the power output of the solar panels to the power output of the nuclear power plant (1 GigaWatt = 10^9 Watts) to determine the required area.
Area of solar panels = Power output of nuclear power plant / Power output of the solar panels
Area of solar panels = (1 GigaWatt) / [(10%)(4*10^33 erg/s) / (10^7 erg/s)]
Now, let's calculate the area:
Area of solar panels = (1 x 10^9 Watts) / [(10%)(4*10^33 erg/s) / (10^7 erg/s)]
Simplifying further:
Area of solar panels = (1 x 10^9) / [(0.10)(4*10^33) / (10^7)]
Area of solar panels = (1 x 10^9) / [4 x 10^34 / 10^7]
Area of solar panels = (1 x 10^9) / (4 x 10^27)
Area of solar panels = (1/4) x (10^9 / 10^27)
Area of solar panels = 2.5 x 10^-19 km^2
Therefore, the solar panels must cover an area of approximately 2.5 x 10^-19 km^2 to match the power output of a large nuclear power plant.
To determine the area required for the solar panels to match the power output of a gigawatt, we need to convert the given luminosity of the Sun to watts.
Given:
Luminosity of the Sun = 4 * 10^33 erg/s
To convert ergs to watts, we'll use the conversion factor:
1 erg/s = 10^(-7) watts
So, the luminosity of the Sun in watts is:
4 * 10^33 erg/s * 10^(-7) watts/erg = 4 * 10^26 watts
Now, let's calculate the power output of the Sun falling on each square meter:
Power output per square meter = Luminosity of the Sun / (4 * π * (radius of the Sun)^2)
= 4 * 10^26 watts / (4 * π * (7 * 10^10 cm)^2)
First, convert the radius of the Sun from cm to meters:
Radius of the Sun = 7 * 10^10 cm * (1 meter / 100 cm) = 7 * 10^8 meters
Now, substitute the values into the formula:
Power output per square meter = 4 * 10^26 watts / (4 * π * (7 * 10^8 meters)^2)
Now, we need to account for the solar panel efficiency of 10%:
Power output per square meter of solar panels = Power output per square meter * Efficiency
= (4 * 10^26 watts / (4 * π * (7 * 10^8 meters)^2)) * 0.1
To match a gigawatt (1 GW) of power output, calculate the number of square meters needed:
Number of square meters = 1 GW / Power output per square meter of solar panels
Now, convert the units from square meters to square kilometers:
Number of square kilometers = Number of square meters * (1 km^2 / 10^6 m^2)
Final Steps:
1. Calculate the power output per square meter of solar panels using the given values.
2. Calculate the number of square meters needed to match 1 GW of power output.
3. Convert the units from square meters to square kilometers to get the final answer.
Calculating the formula steps will yield the area in square kilometers that your solar panels will need to cover in order to match the power output of a gigawatt.
1 Watt = 1 J/s = 10^7 erg/s
1GW = 10^9 J/s
the area of the sun is
4pi*(7*10^5 km)^2 = 6.2*10^12 km^2
So the solar output is
4*10^17 GW/6.2*10^12km^2
= 6.452*10^4 GW/km^2
So, now the question is: how much of that reaches the earth? I guess you need to use the inverse-square law using the distance from the sun.