CHEMISTRY IN FOCUS (LL)-TEXT
CHEMISTRY IN FOCUS (LL)-TEXT
7th Edition
ISBN: 9781337399845
Author: Tro
Publisher: CENGAGE L
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Chapter 9, Problem 44E
Interpretation Introduction

Interpretation:

The energy produced (in kilocalories) in 1 second and the consumption of coal and natural gas (in kilogram) are to be calculated.

Concept introduction:

The power derived from the utilization of chemical or physical resources to provide heat and light or to carry out various processes is known as energy. The SI unit of energy is joule.

Unit conversion is a multiple-step process that is used to convert the unit of measurement of a given quantity. It is determined by multiplicaiton with a conversion factor.

The useful energy derived can be calculated by using the equation:

Total consumed×efficency=useful energy.

Conversion factor megajoules to joule is 106 J1 MJ

Conversion factor calorie to kilocalorie is 1 kcal1000 cal

Expert Solution & Answer
Check Mark

Answer to Problem 44E

Solution:

2.4×105 kcal/sec

2.9×103 MJ/sec

1.0×102 kg/sec 

60 kg/sec

Explanation of Solution

a) The number of kilocalories produce in 1 s.

The energy produced in one second by the power plant is given as:

Energy=1.0 ×103megajoules/second

The kilocalorie produced in one second is calculated as follows:

First, convert megajoule into joule, then joule into calorie, and then calorie into kilocalorie.

Conversion of megajoule into kilocalorie is as follows:

4.18 joule=1 calorie `

E=(1000 MJsec)×(106 J1 MJ)×(1 cal4.18 J)×(1 kcal1000 cal)=2.4×105 kcal/sec

Therefore, the kilocalorie produced in one second is 2.4×105 kcal/sec.

b) The amount of energy it consumes in 1 s if its efficiency is 34%.

The total energy consumed in one second is calculated as follows:

The equation used to calculate the total energy consumed is represented as follows:

Total consumed×efficency=useful energy

The total energy produced in one second is 2.4×105 kcal/sec. (From subpart a)

The efficiency of a person is 34%.

Conversion of percentage into decimal:

Efficiency=34%Efficiency=34100Efficiency=0.34

Substitute the values in the given equation:

2.4×105 kcal/sec=total consumed×0.34(2.4×105 kcal/sec0.34)=total consumed7.1×105 kcal/sec=total consumed

Or

The energy consumed in MJ is calculated as follows:

1000 MJ=total consumed×0.34(1000 MJ0.34)=total consumed2.9×103 MJ/sec=total consumed

Therefore, the energy consumed (in MJ) is 2.9×103 MJ/sec.

c) The consumption of coal in 1 s if the power plant were coal fired.

The consumption of coal (in kilogram) is calculated as follows:

First, the energy consumed is divided by the heat of combustion of coal. Then, it is converted into kilograms.

Conversion of kilocalorie into kilogram is as follows:

The energy consumed is 7.1×105 kcal/sec. (From subpart b).

The heat of combustion of coal is 6.8 kcal. (From table 9.6).

Therefore,

E=(7.1×105 kcal/sec)×(1 g6.8 kcal)×(1 kg1000 g)=1.0×102 kg/sec of coal

Therefore, the consumption of coal (in kilogram) is 1.0×102 kg/sec .

d) The consumption of natural gas in 1 s if the power plant were natural gas fired.

The consumption of natural gas (in kilogram) is calculated as follows:

First, the energy consumed is divided by the heat of combustion of natural gas. Then, it is converted into kilograms.

Conversion of kilocalorie into kilogram is as follows:

The energy consumed is 7.1×105 kcal/sec. (From subpart b).

The heat of combustion of natural gas is 11.8 kcal. (From table 9.6).

Therefore,

E=(7.1×105 kcal/sec)×(1 g11.8 kcal)×(1 kg1000 g)=60 kg/sec of natural gas

Therefore, the consumption of natural gas (in kilogram) is 60 kg/sec.

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Chapter 9 Solutions

CHEMISTRY IN FOCUS (LL)-TEXT

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