Question

The intensity of solar radiation that falls on a detector on Earth is $1.00 \mathrm{kW} / \mathrm{m}^{2} .\( The detector is a square that measures \)5.00 \mathrm{m}$ on a side and the normal to its surface makes an angle of \(30.0^{\circ}\) with respect to the Sun's radiation. How long will it take for the detector to measure \(420 \mathrm{kJ}\) of energy?

Short answer

Answer: It will take the solar detector approximately 19.40 seconds to measure 420 kJ of energy.

Step-by-step solution

Calculate the area of the solar detector.

Given that the detector is a square with each side measuring 5.00 m, we can calculate the area (A) of the detector using the formula for the area of a square: A = side × side A = 5.00 m × 5.00 m A = 25.00 m²

Calculate the projected area of the solar detector.

Since the surface normal of the solar detector makes an angle of 30 degrees with the Sun's radiation, we need to find the projected area (A_proj) of the solar detector in the direction of the radiation. We can use the following formula: A_proj = A × cos(angle) Where angle = 30 degrees A_proj = 25.00 m² × cos(30°) A_proj ≈ 21.65 m²

Calculate the power received by the solar detector.

Given that the intensity (I) of solar radiation is 1.00 kW/m², we can calculate the power (P) received by the solar detector using the formula: P = I × A_proj P = 1.00 kW/m² × 21.65 m² P ≈ 21.65 kW

Calculate the energy received by the solar detector.

We have the energy (E) to be measured by the detector, which is 420 kJ. To find the time (t) it takes to measure this amount of energy, we will use the formula: E = P × t So, t = E / P Where E = 420 kJ

Convert the units of power to be consistent with the energy.

Before we find the time, we need to have consistent units for power and energy. Given that the energy is in kJ, we must convert the power from kW to kJ/s: 1 kW = 1 kJ / s So, P ≈ 21.65 kJ/s

Calculate the time required for the detector to measure the energy.

Now that we have consistent units for energy and power, we can find the time required for the detector to measure 420 kJ of energy: t = E / P t = 420 kJ / 21.65 kJ/s t ≈ 19.40 s Therefore, it will take the solar detector approximately 19.40 seconds to measure 420 kJ of energy.