In a sunny location, sunlight has a power density of about 1 kW/m². Photovoltaic solar cells can convert this power into electricity with 15% efficiency. If a typical home uses 385 kWh of electricity per month, how many square meters of solar cells are required to meet its energy requirements? Assume that electricity can be generated from the sunlight for 8 hours per day.

Verified step by step guidance

Calculate the total energy consumption of the home in kilowatt-hours (kWh) per day by dividing the monthly consumption by the number of days in a month.

Convert the daily energy consumption from kilowatt-hours to kilowatts by dividing by the number of hours in a day that the solar cells can generate electricity.

Determine the amount of power that needs to be generated by the solar cells by dividing the required power by the efficiency of the solar cells (15%).

Calculate the area of solar cells needed by dividing the required power by the power density of sunlight (1 kW/m^2).

Ensure the units are consistent and check the calculations to verify the area of solar cells required.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Power Density

Power density refers to the amount of power (energy per unit time) received per unit area, typically measured in watts per square meter (W/m²). In this context, a power density of 1 kW/m² indicates that each square meter of solar panel can receive 1 kilowatt of solar energy under optimal conditions. Understanding power density is crucial for calculating the total energy available from sunlight for solar energy applications.

Efficiency of Solar Cells

The efficiency of solar cells is the ratio of the electrical output to the solar energy input, expressed as a percentage. In this case, a 15% efficiency means that only 15% of the solar energy hitting the solar cells is converted into usable electricity. This concept is vital for determining how much solar panel area is needed to generate a specific amount of electricity, as higher efficiency results in less area required.

Energy Consumption and Conversion

Energy consumption refers to the total amount of energy used by a household, measured in kilowatt-hours (kWh). To meet the energy needs of a home using 385 kWh per month, one must convert this monthly requirement into a daily average and then into the equivalent solar energy needed, factoring in the hours of sunlight available. This conversion is essential for calculating the area of solar cells required to meet the household's energy demands.

Recommended video:

Guided course

01:56

Conversion Factors

Related Practice

Textbook Question

Determine the mass of CO₂ produced by burning enough of each fuel to produce 1.00×10² kJ of heat. a. CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g) ΔH°_rxn = –802.3 kJ

views

Textbook Question

Methanol (CH₃OH) has been suggested as a fuel to replace gasoline. Find ΔH°_rxn, and determine the mass of carbon dioxide emitted per kJ of heat produced. Use the information from the previous exercise to calculate the same quantity for octane, C₈H₁₈. How does methanol compare to octane with respect to global warming?

Textbook Question

The kinetic energy of a rolling billiard ball is given by KE = 1/2 mv². Suppose a 0.17-kg billiard ball is rolling down a pool table with an initial speed of 4.5 m/s. As it travels, it loses some of its energy as heat. The ball slows down to 3.8 m/s and then collides head-on with a second billiard ball of equal mass. The first billiard ball completely stops and the second one rolls away with a velocity of 3.8 m/s. Assume the first billiard ball is the system. Calculate q.

views

Textbook Question

LP gas burns according to the exothermic reaction: C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g) ΔH°_rxn = –2044 kJ What mass of LP gas is necessary to heat 1.5 L of water from room temperature (25.0 °C) to boiling (100.0 °C)? Assume that during heating, 15% of the heat emitted by the LP gas combustion goes to heat the water. The rest is lost as heat to the surroundings.