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A rooftop in the southwestern United States receives an average solar power of 3 .0 × 10 5 W (averaged over both day and night and seasons). Convert this to kilowatts and then multiply by the number of hours in a year to calculate the number of kilowatt-hours that fall on the rooftop in one year. If the home uses 1 .0 × 10 3 kilowatt-hours per month, does its roof receive enough energy to meet its energy needs? What is the minimum efficiency required for solar cells to meet the entire energy need?

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Chemistry In Focus

7th Edition
Tro + 1 other
Publisher: Cengage Learning,
ISBN: 9781337399692

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Chapter
Section
BuyFindarrow_forward

Chemistry In Focus

7th Edition
Tro + 1 other
Publisher: Cengage Learning,
ISBN: 9781337399692
Chapter 10, Problem 32E
Textbook Problem
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A rooftop in the southwestern United States receives an average solar power of 3 .0 × 10 5 W (averaged over both day and night and seasons). Convert this to kilowatts and then multiply by the number of hours in a year to calculate the number of kilowatt-hours that fall on the rooftop in one year. If the home uses 1 .0 × 10 3 kilowatt-hours per month, does its roof receive enough energy to meet its energy needs? What is the minimum efficiency required for solar cells to meet the entire energy need?

Interpretation Introduction

Interpretation:

The average solar power in kilowatts and the number of kilowatt-hours that fall on the rooftop in one year are to be calculated and whether the roof receives enough energy to meet the energy needs or not and the minimum percent efficiency of the solar cells are to be determined.

Concept Introduction:

Power input on a PV cell is the total active area of the PV cell multiplied by the solar power.

The percent of the input power that is converted into the output power by a PV cell is called its efficiency.

Percent efficiency of a PV cell is 100 multiplied by the ratio of power out and power in.

% efficiency=power outpower in×100% …… (1)

Explanation of Solution

Given information: The rooftop receives an average solar power of 3.0 × 105 W.

The energy output required per month is 1.0 × 103kWh.

The relation between watt and kilowatt is given as follows:

1000 W=1 kW

Convert 3.0 × 105 W into kilowatt by using the above relation as:

3.0 × 105 W=(3.0 × 105 W×1 kW1000 W)=300 kW

The relation between years and days is as:

1 year=365 days

The relation between days and hours is as:

1 day=24 hours

Convert 300 kW into kWh with the help of the relations given above as:

300 kW=(300 kW×365 days1 yr×24 h1 day)=2

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Chemistry In Focus
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