   # The best solar panels currently available are about 15% efficient in converting sunlight to electricity. A typical home will use about 40. kWh of electricity per day (1 kWh = 1 kilowatt hour; 1kW = 1000 J/s). Assuming 8.0 hours of useful sunlight per day, calculate the minimum solar panel surface area necessary to provide all of a typical home’s electricity. (See Exercise 124 for the energy rate supplied by the sun.) ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 7, Problem 125CP
Textbook Problem
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## The best solar panels currently available are about 15% efficient in converting sunlight to electricity. A typical home will use about 40. kWh of electricity per day (1 kWh = 1 kilowatt hour; 1kW = 1000 J/s). Assuming 8.0 hours of useful sunlight per day, calculate the minimum solar panel surface area necessary to provide all of a typical home’s electricity. (See Exercise 124 for the energy rate supplied by the sun.)

Interpretation Introduction

Interpretation: The surface area of solar panel should be calculated.

### Explanation of Solution

Explanation

Record data from given:

Sun furnished energy is 1.0 kW/m2

Efficiency of solar panel is 1500

Electricity per used day is 40. kWh

The useful sunlight per day is 8h

To calculated the solar panel utilized sun energy per day.

1 kW = 1kJ/s40kWh=40kJh=40.kJhs×3600sh=1.4×105kJ

• A typical home used electricity per day is 8 hours so the given 40. kWh is multiplied by the 3600 s per hour to get the sun energy used in 8 hours.
• Sun energy utilized in solar panel for 8 hour is 1.4×105kJ .

To calculate the furnished sun energy for 8 hours,

Furnishedsunenergy=1.0kWsm21kWsm2=1kJsm2=1

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