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Heating the air in a particular house by 10°C requires about 5 .0 × 10 3 kJ of heat energy. By assuming the heat came from natural gas with 80% efficiency, how many grams of natural gas are required? By assuming the heat came from electricity with 80% efficiency, and the electricity was produced from the combustion of coal with 30% efficiency, how much coal would be required?

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Chemistry In Focus

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

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Section
BuyFindarrow_forward

Chemistry In Focus

7th Edition
Tro + 1 other
Publisher: Cengage Learning,
ISBN: 9781337399692
Chapter 9, Problem 45E
Textbook Problem
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Heating the air in a particular house by 10°C requires about 5 .0 × 10 3 kJ of heat energy.

By assuming the heat came from natural gas with 80% efficiency, how many grams of natural gas are required?

By assuming the heat came from electricity with 80% efficiency, and the electricity was produced from the combustion of coal with 30% efficiency, how much coal would be required?

Interpretation Introduction

Interpretation:

The amount of natural gas (in grams) and coal (in grams) required 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 multiplication with a conversion factor.

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

Total consumed×efficency=useful energy.

Explanation of Solution

a) The number of grams of natural gas are required when the heat came from gas with 80% efficiency.

The amount of natural gas required is calculated as follows:

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

Total consumed×efficency=useful energy

The total energy required is 5.0×103 kJ.

The efficiency of natural gas is 80%.

Conversion of percentage into decimal:

Efficiency=80%Efficiency=80100Efficiency=0.80

Substitute the values in the given equation:

5.0×103 kJ=total consumed×0.80(5.0×103 kJ0.80)=total consumed6.25×103 kJ=total consumed

Therefore, the energy consumed is 6.25×103 kJ.

Now, the energy consumed is divided by the heat of combustion of natural gas.

The amount of natural gas required is as follows:

The heat of combustion of natural gas is 49.3 kJ. (From table 9.6).

E=(6.25×103 kJ)×(1 g49.3 kJ)=1.3×102 g

Therefore, amount of natural gas required is 1.3×102 g.

b) The amount of coal required when the heat came from electricity with 80% efficiency and the electricity was produced from the combustion of coal with 30% efficiency.

The amount of coal required is calculated as follows:

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

Total consumed×efficency=useful energy

The total energy required is 5.0×103 kJ.

The efficiency is 80%.

Conversion of percentage into decimal:

Efficiency=80%Efficiency=80100Efficiency=0

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Chemistry In Focus
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