How much heat must be applied to the different gases in the atmosphere to increase its temperature by 1°C?

 

Author; Rogelio Pérez C

Abstract

The greenhouse theory teaches that the increase in temperature on the planet is caused by heat-trapping gases called ghg such as CO2, so to lower the temperature of the atmosphere it is necessary to decrease these gases in the atmosphere, but these claims are contrary to the specific heat capacity of greenhouse gas  such as CO2, This work shows that GHGS due to the ability to trap heat require a greater amount of heat, in order to increase their temperature by 1°C, i.e. the main greenhouse gases (CO2.Water Vapor, CH4) reduce the temperature by about 1/3 of the heat they absorb. So it can be concluded that, contrary to the theory of the greenhouse effect, it is necessary to increase the ghg in the atmosphere to reduce the temperature of the planet by almost 1/3. Experimentally we can see it in industry, with the technique known as oxy combustion, where it uses oxygen gas to increase the temperature of an oven and uses CO2 gas to reduce the heat of the oven. Also in the cities of the world, water vapor is greater in the coldest cities of the planet, and the warmer cities have lower concentration of water vapor. Example; average annual humidity (water vapor) from Ottawa, Canada is 80%, and temperatures between -14°C and 14°C, Riyadh, Saudi Arabia with an average humidity (water vapor)  of 30%, and average temperatures of 32°C.

Introduction

the atmosphere is the gaseous body that surrounds the planet transferring its temperature to us, the greenhouse effect teaches us that the temperature of the atmosphere is caused only by certain gases, called greenhouse gases, which have a volume of 0.04% of the gases of the atmosphere, which because they retain heat from the surface of the earth, and they emit it again in all directions at higher temperature, this work based on the specific heat of the gases of the atmosphere, shows how much heat each gas in the atmosphere needs, to raise its temperature by 1° C.

 

Scientific theory:

The equipartition theorem relates the temperature of a system to its average energies. It makes quantitative predictions, provides the total kinetic and potential energies for a system at a given temperature, from which the heat capacity of the system can be calculated. However, the equipartition also provides the average values of individual energy components, such as the kinetic energy of a particular particle or the potential energy of a single spring. For example, it predicts that each atom in an ideal monoatomic gas has an average kinetic energy of (3/2) k B T in thermal equilibrium, where k B is Boltzmann's constant and Te the temperature (thermodynamics).1

The greenhouse effect is a process in which thermal radiation emitted by the planetary surface is absorbed by atmospheric greenhouse gases (GHGs) and radiated in all directions. As part of this radiation is returned to the Earth's surface and lower atmosphere, resulting in an increase in the average surface temperature compared to what would be in the absence of GHGs.2

One part of the solar radiation that reaches Earth passes through the atmosphere, is reflected back into space; another reaches the ground and warms it. It emits heat (infrared radiation) and warms the atmosphere, since the heat is retained by the greenhouse gases.3

Without this natural greenhouse effect, the equilibrium temperature of the Earth would be about -18 °C. However, the average temperature of the earth's surface is about 14 °C, a difference close to 32 °C that gives us an idea of the magnitude of efecto.4

Temperature is a measure of the average kinetic energy of the atoms or molecules in the system.5

Heat, is thermal energy transferred from a hotter system to a cooler system that are in contact.

We can calculate the heat released or absorbed using the specific heat capacity C, end the mass of the substance m, and the change in temperature ΔT, end text in the equation:  Q= M×C×ΔT.  6

Specific Heat: The amount of heat needed to raise the temperature of 1kg of a substance by 1°C.7

 

In industry CO2 as a heat reducer

Oxy-combustion; it is a technique that consists in separating the nitrogen of the atmosphere of an oven, and replace it with pure oxygen, which increases the temperature. When the temperature is 3000°C by the injection of pure oxygen, then is recirculated the CO2 produced by the oven to reduce the temperature, which lowers the temperature of the oven until the 1900°C.8

 

Results;

Physics teaches us that; "Heat is thermal energy transferred from a hotter system to a cooler system that is in contact," and "We can calculate the heat released or absorbed using the specific heat capacity, the mass of the substance, and the temperature change in the equation": Q=M×C×ΔT.

Q = heat released or absorbed

m = mass of the substance

C = Specific heat; this is the amount of heat needed by one gram of a substance to raise its temperature by 1°C.

ΔT= Temperature change.

 

1-   How much heat should be applied to 0.028 kg of Nitrogen, so that it raises its temperature from 287 K to 288 K?

 

C=Specific heat (N2)=1040 J/Kg°C

Ti= Initial temperature=287k or 14°C

Tf= Final temperature=288k or 15°C

M= mass of N2= 0.028 Kg

Q=?

Equation; Q= M × C × ΔT.

 

Solution; Q= (0.028 Kg) x (1040 J/Kg°C) x (288 k or15°C- 287 k or 14°C)

  Q= (0.028 Kg) x (1040 J/Kg°C) x (1°C)

  Q= (0.028) x (1040 J) x (1)

 Q= 29.12 J

The amount of heat needed for N2 to rise from 287 kelvin or 14°C to 288 kelvin or 15°C is 29.12 Joules.

Now we will convert joules into kelvin degrees based on the following;

Of the two expressions for the pressure of a gas, one derived from macroscopic experimental data, P · V = k · T and another derivative of Newton's laws, P = m * (v2) pr / 3V. If both describe the same reality, then it should happen that k · T = m * (v2) pr / 3. It follows that the temperature, T = 2 / (3K) * m * (v2) pr / 2, that is, the temperature of a gas is proportional to the average kinetic energy of its molecules.

T = 2 / (3k) · m · (v2) pr / 2

T = 2 / (3R 8, 31 J / mole .k) Kinetic Energy = 29.12 J

T = (0,0802246289610911J / mole .k). 29.12 J

T = (0, 0802246289610911 k). 29.12

T = 2.33 Kelvin

29, 12 joules equals 2, 33 kelvin of temperature, then we can say that for every 2.33 k (29.12 J) of heat absorbed by the N2 in the atmosphere, it increases its temperature by 1K or 1°C., a decrease of more or less 1/2 of the temperature.

2-   How much heat should be applied to 0.032 kg of Oxygen, so that it raises its temperature from 287 K to 288 K?

 

C=Specific heat (O2) =920 J/Kg°C

Ti= Initial temperature=287k or 14°C

Tf= Final temperature=288k or 15°C

M= mass of O2= 0.032 Kg

Q=?

Equation; Q= M × C × ΔT.

Solution; Q= (0.032 Kg) x (920 J/Kg°C) x (288 k or15°C- 287 k or 14°C)

  Q= (0.032 Kg) x (920 J/Kg°C) x (1°C)

  Q= (0.032) x (920 J) x (1)

 Q= 29.44 J

The amount of heat needed for O2 to rise from 287 kelvin or 14°C to 288 kelvin or 15°C is 29.44 Joules.

Now we will convert joules into kelvin degrees based on the following Equation;

T = 2 / (3k) · m · (v2) pr / 2

T = 2 / (3R 8, 31 J / mole .k) Kinetic Energy = 29.44 J

T = (0,0802246289610911J / mole .k). 29.44 J

T = (0, 0802246289610911 k). 29.44

T = 2.36 Kelvin

29, 12 joules equals 2, 36 kelvin of temperature, then we can say that for every 2.36 k (29.44 J) of heat absorbed by the O2 in the atmosphere, it increases its temperature by 1K or 1°C a decrease of more or less 1/2 of the temperature.

3-   Now much heat should be applied to 0.040 kg of Argon, so that it raises its temperature from 287 K to 288 K?

 

C=Specific heat (Ar)=310 J/Kg°C

Ti= Initial temperature=287k or 14°C

Tf= Final temperature=288k or 15°C

M= mass of N2= 0.040 Kg

Q=?

Equation; Q= M × C × ΔT.

 

Solution; Q= (0.040 Kg) x (310 J/Kg°C) x (288 k or15°C- 287 k or 14°C)

  Q= (0.040 Kg) x (310 J/Kg°C) x (1°C)

  Q= (0.040) x (310 J) x (1)

 Q= 12.4 J

The amount of heat needed for Ar to rise from 287 kelvin or 14°C to 288 kelvin or 15°C is 12.4 Joules.

Now we will convert joules into kelvin degrees based on the following Equation;

T = 2 / (3k) · m · (v2) pr / 2

T = 2 / (3R 8, 31 J / mole .k) Kinetic Energy = 12.4 J

T = (0,0802246289610911J / mole .k). 12.4 J

T = (0, 0802246289610911 k). 12.4

T = 0.99 Kelvin

12.4 joules equals 0.99 kelvin of temperature, and then we can say that for every 0.99 K (12.4 J) of heat absorbed by the Ar in the atmosphere, it increases its temperature by 1K or 1°C.a increase of 0.1 in temperature.

 

4-    Now much heat should be applied to 0.044 kg of CO2, so that it raises its temperature from 287 K to 288 K?

 

C=Specific heat (CO2)=836 J/Kg°C

Ti= Initial temperature=287k or 14°C

Tf= Final temperature=288k or 15°C

M= mass of CO2= 0.044 Kg

Q=?

Equation; Q=m×C×ΔT.

Solution; Q= (0.044 Kg) x (836 J/Kg°C) x (288 k or15°C- 287 k or 14°C)

  Q= (0.044 Kg) x (836 J/Kg°C) x (1°C)

  Q= (0.044) x (836 J) x (1)

 Q= 36.784 J

The amount of heat needed for CO2 to rise from 287 kelvin or 14°C to 288 kelvin or 15°C is 36.794 Joules.

Now we will convert joules into kelvin degrees based on the following Equation;

T = 2 / (3k) · m · (v2) pr / 2

T = 2 / (3R 8, 31 J / mole .k) Kinetic Energy = 36.784 J

T = (0,0802246289610911J / mole .k). 36.784 J

T = (0, 0802246289610911 k). 36.784

T = 2.95 Kelvin

36.784 joules equals 2, 95 kelvin of temperature, then we can say that for every 2.95 k (36.784 J) of heat absorbed by the CO2 in the atmosphere, it increases its temperature by 1K or 1°C., a decrease of about 1/3 of the temperature.

 

5-   How much heat should be applied to 0.018 kg of Water vapor, so that it raises its temperature from 287 K to 288 K?

 

C=Specific heat (O2) =2016 J/Kg°C

Ti= Initial temperature=287k or 14°C

Tf= Final temperature=288k or 15°C

M= mass of O2= 0.018 Kg

Q=?

Equation; Q= M × C × ΔT.

Solution; Q= (0.018 Kg) x (2016 J/Kg°C) x (288 k or15°C- 287 k or 14°C)

  Q= (0.018 Kg) x (2016 J/Kg°C) x (1°C)

  Q= (0.018) x (2016 J) x (1)

 Q= 36.28 J

The amount of heat needed for Water vapor to rise from 287 kelvin or 14°C to 288 kelvin or 15°C is 36.28 Joules.

Now we will convert joules into kelvin degrees based on the following Equation;

T = 2 / (3k) · m · (v2) pr / 2

T = 2 / (3R 8, 31 J / mole .k) Kinetic Energy = 36.28 J

T = (0,0802246289610911J / mole .k). 36.28 J

T = (0, 0802246289610911 k). 36.28

T = 2.91 Kelvin

36.28 joules equals 2.91 kelvin of temperature, and then we can say that for every 2.91 k (36.28 J) of heat absorbed by the Water vapor in the atmosphere, it increases its temperature by 1K or 1°C., a decrease of about 1/3 of the temperature.

6-   Now much heat should be applied to 0.0164 kg of Methane, so that it raises its temperature from 287 K to 288 K?

 

C=Specific heat (CH4) =2200 J/Kg°C

Ti= Initial temperature=287k or 14°C

Tf= Final temperature=288k or 15°C

M= mass of CH4= 0.0164 Kg

Q=?

Equation; Q=M×C×ΔT.

Solution; Q= (0.0164 Kg) x (2200 J/Kg°C) x (288 k or15°C- 287 k or 14°C)

  Q= (0.0164 Kg) x (2200 J/Kg°C) x (1°C)

  Q= (0.0164) x (2200 J) x (1)

 Q= 36.08 J

The amount of heat needed for CO2 to rise from 287 kelvin or 14°C to 288 kelvin or 15°C is 36.08 Joules.

Now we will convert joules into kelvin degrees based on the following Equation;

T = 2 / (3k) · m · (v2) pr / 2

T = 2 / (3R 8, 31 J / mole .k) Kinetic Energy = 36.08 J

T = (0,0802246289610911J / mole .k). 36.08 J

T = (0, 0802246289610911 k). 36.08

T = 2.89 Kelvin

36.08 joules equals 2, 89 kelvin of temperature, then we can say that for every 2.89 k (36.08 J) of heat absorbed by the CH4 in the atmosphere, it increases its temperature by 1K or 1°C., a decrease of about 1/3 of the temperature.

Conclusion;

Based on the specific heat of the different gases that make up the atmosphere, the ghgs of the atmosphere must be applied more heat than the other gases of the atmosphere to raise their temperature by 1°C, Because of its nature to catch more heat. Therefore, it can be concluded that ghgs in the atmosphere reduce the temperature by almost 1/3, Contrary to the explanation of the theory of the greenhouse effect, this process of decreasing the temperature by gases such as CO2 and water vapor, we can see it experimentally in industry and in large cities, where in the industry a technique called oxy combustion is used, where the temperature of an oven is increased with the oxygen gas, and decreased with the co2 gas. Also in the cities of the world, water vapor is greater in the coldest cities of the planet, and the warmer cities have lower concentration of water vapor. Example; average annual water vapor from Ottawa, Canada is 80%, and temperatures between -14°C and 14°C, Riyadh, Saudi Arabia with an average humidity of 30%, and average temperatures of 32°C.

Bibliography

1-  http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/eqpar.html

2-  Intergovernmental Panel on Climate Change.

3-  Taking the Earth’s Temperature». American Chemical Society. Atmospheres (en inglés) 118 (8): 3213-3217. ISSN 2169-8996. doi:10.1002/jgrd.50359.

4-  Jones, P. D.; Harpham, C. (2013). «Estimation of the absolute surface air temperature of the Earth». Journal of Geophysical Research. CHAPTER 7. THE GREENHOUSE EFFECT». acmg.seas.harvard.edu.

5-  https://www.khanacademy.org/science/chemistry/thermodynamics-chemistry/internal-energy-sal/a/heat

6-  https://www.khanacademy.org/science/chemistry/thermodynamics-chemistry/internal-energy-sal/a/heat

7-https://www.thoughtco.com/definition-of-specific-heat-capacity-605672

8-http://ainenergia.com/oxicombustion-en-centrales-termicas-renovarse-o-morir/