In a large building, oil is used in a steam boiler heating system. The combustion of 1.0 lb of oil provides 2.4 × 10 7 J. b. How many kilograms of oil are needed to change 150 kg of water to steam at 100 °C?

Hello. In this problem, we are told that the gas a byproduct in the sugar cane milling process can be used as a fuel in steam production. If the combustion of £2 of a gas releases 1.95 times 10 to 7 jews of heat, calculate the amount of a gas in kilograms needed to convert 275 kg of water to steam. We are provided the anthropology of vaporization of water as 2.26 times 10 to third jewels per gram. From the second sense, we can relate the heat to the mass of the gas. We have 1.95 times 10 to the seven jewels divided by £2 of the gas. And from the NTH P vaporization, we have 2.26 times 10 to the third jewels for every one g of water. Within this problem, we have mass and kilograms and pounds. So we can make use of a couple of conversion factors to go from one unit to the other. We call that £1 is equal to 454 g and one kg is equal to 1000 g. We'll begin our calculation with the mass of water that we need to convert to steam. This is 275 kg of water. We will convert our mass from kilograms to grams by multiplying by g and divided by one kg are kilograms of water in the numerator. And denominator cancels, we can then make use of the entropy of vaporization to go from mass to energy. We'll multiply by 2.26 have 10, that third jewels. And if I buy one g of water, our mass of water in the numerator and denominator cancels, you then can relate energy to the mass of the gas. We have £2 of a gas we multiply by and then we divide by 1.95 times 10 to 7 jewels are units of energy and jewels. Numerator and denominator cancels, we need to get our mass from pounds two kg. So we will begin by going to grams. So we multiply by 454 g and divide by £1 our pounds of the gas and the numerator and denominator cancel. And then we'll convert our mass from grams to kilograms by multiplying by one kg and dividing by 1000 g, our grams of gas in the numerator and denominator cancel. We left with kilograms of gas and this is equal to 28. kilograms of the gas. So this is the amount of gas needed to convert 275 kg of water to steam. Thanks for watching. Hope this helped.

Table of contents

Skip topic navigation

1. Matter and Measurements4h 31m
2. Atoms and the Periodic Table5h 23m
3. Ionic Compounds2h 18m
4. Molecular Compounds2h 18m
5. Classification & Balancing of Chemical Reactions3h 17m
6. Chemical Reactions & Quantities2h 27m
7. Energy, Rate and Equilibrium3h 46m
8. Gases, Liquids and Solids3h 25m
9. Solutions4h 10m
10. Acids and Bases3h 29m
11. Nuclear Chemistry56m
BONUS: Lab Techniques and Procedures1h 38m
BONUS: Mathematical Operations and Functions47m
12. Introduction to Organic Chemistry1h 34m
13. Alkenes, Alkynes, and Aromatic Compounds2h 12m
14. Compounds with Oxygen or Sulfur1h 6m
15. Aldehydes and Ketones1h 1m
16. Carboxylic Acids and Their Derivatives1h 11m
17. Amines39m
18. Amino Acids and Proteins1h 51m
19. Enzymes1h 37m
20. Carbohydrates1h 41m
21. The Generation of Biochemical Energy2h 8m
22. Carbohydrate Metabolism2h 22m
23. Lipids2h 26m
24. Lipid Metabolism1h 45m
25. Protein and Amino Acid Metabolism1h 37m
26. Nucleic Acids and Protein Synthesis2h 54m

8. Gases, Liquids and Solids

Heating and Cooling Curves

GOB Chemistry

8. Gases, Liquids and Solids

Heating and Cooling Curves

Previous problemNext problem

2:53 minutes

Problem 99b

Textbook Question

In a large building, oil is used in a steam boiler heating system. The combustion of 1.0 lb of oil provides 2.4 × 10⁷ J.
b. How many kilograms of oil are needed to change 150 kg of water to steam at 100 °C?

Verified step by step guidance

Determine the amount of energy required to convert 150 kg of water to steam at 100 °C. Use the formula for heat energy:

q = m × L

_v, where

m

is the mass of water (150 kg) and

L

_v is the heat of vaporization of water (2.26 × 10⁶ J/kg).

Calculate the total energy required by substituting the values into the formula:

q = 150 × 2.26 × 10⁶

. This will give the energy in joules needed to convert the water to steam.

Determine how much energy is provided by 1.0 lb of oil. From the problem, 1.0 lb of oil provides 2.4 × 10⁷ J. Convert this energy to the equivalent in kilograms of oil by using the conversion factor: 1 lb = 0.453592 kg.

Calculate the mass of oil required to provide the total energy needed. Use the relationship:

m

_oil = q / E_oil, where

q

is the total energy required (from step 2) and

E

_oil is the energy provided per kilogram of oil (from step 3).

Perform the division to find the mass of oil in kilograms. This will give the amount of oil needed to convert 150 kg of water to steam at 100 °C.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above

Video duration:

Play a video:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Heat of Vaporization

The heat of vaporization is the amount of energy required to convert a unit mass of a substance from liquid to gas at its boiling point. For water, this value is approximately 2.26 x 10⁶ J/kg. Understanding this concept is crucial for calculating how much energy is needed to convert 150 kg of water to steam.

Energy Conversion

Energy conversion refers to the process of changing energy from one form to another. In this context, the energy produced from burning oil is converted into thermal energy to heat water. Knowing the energy content of the oil (2.4 x 10⁷ J per pound) allows us to determine how much oil is needed to provide the required energy for vaporizing the water.

Mass-Energy Relationship

The mass-energy relationship is a principle that relates mass to energy, often expressed through the equation E=mc². In practical terms, this concept helps in understanding how the mass of oil needed can be calculated based on the energy required to heat and vaporize water. This relationship is essential for converting the energy requirements into a mass of oil needed for the heating process.

Watch next

Master The Heating Curve with a bite sized video explanation from Jules

Start learning

Related Videos

Related Practice

Textbook Question

The following graph is a heating curve for chloroform, a solvent for fats, oils, and waxes:

d. At the following temperatures, is chloroform a solid, liquid, or gas?

–80 °C, –40 °C; 25 °C; 80 °C

619

views

Textbook Question

What is a liquid's heat of vaporization?

2228

views

Textbook Question

A hot-water bottle for a patient contains 725 g of water at 65 °C. If the water cools to body temperature (37 °C), how many kilojoules of heat could be transferred to sore muscles?

877

views

Textbook Question

A 45-g piece of ice at 0.0 °C is added to a sample of water at 8.0 °C. All of the ice melts and the temperature of the water decreases to 0.0 °C. How many grams of water were in the sample?

1231

views

Textbook Question

In a large building, oil is used in a steam boiler heating system. The combustion of 1.0 lb of oil provides 2.4 × 10⁷ J.

a. How many kilograms of oil are needed to heat 150 kg of water from 22 °C to 100 °C?

843

views

Multiple Choice

If 53.2kJ of heat are added to a 15.5g ice cube at - 5.00 ^oC, what will be the resulting state and temperature of the substance?

Assume that you have a sample of gas at 350 K in a sealed container, as represented in part (a). Which of the drawings (b)–(d) represents the gas after the temperature is lowered from 350 K to 150 K and if the gas has a boiling point of 200 K? Which drawing represents the gas at 150 K if the gas has a boiling point of 100 K?

<IMAGE>

1360

views

Textbook Question

Using the values for the heat of fusion, specific heat of water, and/or heat of vaporization, calculate the amount of heat energy in each of the following:

c. kilojoules needed to melt 24.0 g of ice at 0 °C, warm the liquid to 100 °C, and change it to steam at 100 °C

1602

views

Show more practices