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:

$\text{Total consumed}\times \text{efficency}=\text{useful energy}$.

Conversion factor megajoules to joule is $\frac{{10}^6\text{ J}}{1\text{ MJ}}$

Conversion factor calorie to kilocalorie is $\frac{1\text{ kcal}}{1000\text{ cal}}$

Answer to Problem 44E

Solution:

$2.4\times {10}^5\text{ kcal/sec}$

$2.9\times {10}^3\text{ MJ/sec}$

$1.0\times {10}^2\text{ kg/sec }$

$60\text{ kg/sec}$

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:

$\text{Energy}=\text{1}.0\times \text{1}{0}^{\text{3}}\text{megajoules}/\text{second}$

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:

$4.18\text{ joule}=1\text{ calorie}$ `

$\begin{array}{c}E=\left(\frac{1000\cancel{\text{MJ}}}{\text{sec}}\right)\times \left(\frac{{10}^6\cancel{\text{J}}}{1\cancel{\text{MJ}}}\right)\times \left(\frac{1\cancel{\text{cal}}}{\text{4}\text{.18 }\cancel{\text{J}}}\right)\times \left(\frac{1\text{ kcal}}{1000\cancel{\text{cal}}}\right)\\ =2.4\times {10}^5\text{ kcal/sec}\end{array}$

Therefore, the kilocalorie produced in one second is $2.4\times {10}^5\text{ kcal/sec}$.

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:

$\text{Total consumed}\times \text{efficency}=\text{useful energy}$

The total energy produced in one second is $2.4\times {10}^5\text{ kcal/sec}$. (From subpart a)

The efficiency of a person is $34\%$.

Conversion of percentage into decimal:

$\begin{array}{l}\text{Efficiency}=34\%\\ \text{Efficiency}=\frac{34}{100}\\ \text{Efficiency}=0.34\end{array}$

Substitute the values in the given equation:

$\begin{array}{c}\text{2}\text{.4}\times {\text{10}}^5\text{ kcal/sec}=\text{total consumed}\times \text{0}\text{.34}\\ \left(\frac{\text{2}\text{.4}\times {\text{10}}^5\text{ kcal/sec}}{\text{0}\text{.34}}\right)=\text{total consumed}\\ 7.1\times {10}^5\text{ kcal/sec}=\text{total consumed}\end{array}$

Or

The energy consumed in MJ is calculated as follows:

$\begin{array}{c}1000\text{ MJ}=\text{total consumed}\times \text{0}\text{.34}\\ \left(\frac{1000\text{ MJ}}{\text{0}\text{.34}}\right)=\text{total consumed}\\ 2.9\times {10}^3\text{ MJ/sec}=\text{total consumed}\end{array}$

Therefore, the energy consumed (in MJ) is $2.9\times {10}^3\text{ MJ/sec}$.

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 $7.1\times {10}^5\text{ kcal/sec}$. (From subpart b).

The heat of combustion of coal is $-6.8\text{ kcal}$. (From table $9.6$).

Therefore,

$\begin{array}{c}E=\left(-7.1\times {10}^5\cancel{\text{kcal}}\text{/sec}\right)\times \left(\frac{1\cancel{\text{g}}}{-6.8\cancel{\text{kcal}}}\right)\times \left(\frac{1\text{ kg}}{1000\cancel{\text{g}}}\right)\\ =1.0\times {10}^2\text{ kg/sec of coal}\end{array}$

Therefore, the consumption of coal (in kilogram) is $1.0\times {10}^2\text{ kg/sec }$.

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 $7.1\times {10}^5\text{ kcal/sec}$. (From subpart b).

The heat of combustion of natural gas is $-11.8\text{ kcal}$. (From table $9.6$).

Therefore,

$\begin{array}{c}E=\left(-7.1\times {10}^5\cancel{\text{kcal}}\text{/sec}\right)\times \left(\frac{1\cancel{\text{g}}}{-11.8\cancel{\text{kcal}}}\right)\times \left(\frac{1\text{ kg}}{1000\cancel{\text{g}}}\right)\\ =60\text{ kg/sec of natural gas}\end{array}$

Therefore, the consumption of natural gas (in kilogram) is $60\text{ kg/sec}$.