Temperature and heat

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The temperature of the soil depends on its heat capacity and the heat transfer to or from the soil surface. The heat capacity is influenced by the soil's composition and the amount of water and water vapor it contains.

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Mechanisms

Iframe

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Code
Concept

Mechanisms

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Microscopic heat

Description

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Heat is nothing more than energy at a microscopic level.

In the case of a gas, it corresponds primarily to the kinetic energy of its molecules.

In liquids and solids, we must take into account the attraction between atoms, which is where potential energy comes into play. Therefore, in these cases, heat corresponds to the energy that particles have and with which they oscillate around the equilibrium point defined by the surrounding particles.

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Temperature

Description

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Temperature is the parameter we use to measure the heat energy contained in an object. Since heat energy can never be negative, it is essential to work with the Kelvin scale, where its zero point corresponds to the complete absence of this energy.

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Heat

Description

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Heat is associated with elements like fire, which causes the temperature of water to rise. The process of heating generates movement, indicating that heat is related to mechanical energy. Even the handle of a pot gets heated, and our bodies are capable of perceiving that temperature. Moreover, fire emits radiation that heats up objects that are exposed to it.

From this, we can infer that by transferring heat, we can raise the temperature of an object, and that the generation of movement is associated with energy.

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Model

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Parameters

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$C$
C
Heat capacity
$M$
M
Mass
kg
$c$
c
Specific heat
J/kg K

Variables

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$Q_f$
Q_f
Final heat
J
$\Delta Q$
DQ
Heat difference
J
$\Delta Q_s$
DQ_s
Heat supplied to liquid or solid
J
$Q_i$
Q_i
Initial heat
J
$\Delta T$
DT
Temperature difference
K
$T_f$
T_f
Temperature in final state
K
$T_i$
T_i
Temperature in initial state
K
$\Delta T_s$
DT_s
Temperature variation in a liquid or solid
K

Calculations


First, select the equation: to , then, select the variable: to

Calculations

Symbol
Equation
Solved
Translated

Calculations

Symbol
Equation
Solved
Translated

Variable Given Calculate Target : Equation To be used




Equations

#
Equation

$ c =\displaystyle\frac{ \displaystyle\sum_i c_i M_i }{ \displaystyle\sum_i M_i }$

c = @SUM( c_i * M_i , i )/@SUM( M_i , i )


$ c =\displaystyle\frac{ C }{ M }$

c = C / M


$ c = \displaystyle\frac{ g_a c_a + g_i c_i + g_c c_c + \theta_w c_w }{1+ \theta_w }$

c =( g_a * c_a + g_i * c_i + g_c * c_c + theta_w * c_w )/(1+ theta_w )


$ \Delta Q = C \Delta T $

DQ = C * DT


$ \Delta Q = M c \Delta T$

DQ = M * c * DT


$ \Delta Q = Q_f - Q_i $

DQ = Q_f - Q_i


$ \Delta T = T_f- T_i$

DT = T_f - T_i

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Heat difference

Equation

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If a body initially has an amount of heat the initial heat ($Q_i$) and subsequently has an amount of heat the final heat ($Q_f$) ($Q_f > Q_i$), heat has been transferred to the body the heat difference ($\Delta Q$). Conversely, if ($Q_f < Q_i$), the body has released heat.

$ \Delta Q = Q_f - Q_i $

$Q_f$
Final heat
$J$
9839
$\Delta Q$
Heat difference
$J$
9840
$Q_i$
Initial heat
$J$
9838

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Temperature Difference (Kelvin)

Equation

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If a system is initially at ($$) and then is at the temperature in final state ($T_f$), the difference will be:

$ \Delta T = T_f- T_i$

$\Delta T$
Temperature difference
$K$
6064
$T_f$
Temperature in final state
$K$
5237
$T_i$
Temperature in initial state
$K$
5236



The temperature difference is independent of whether these values are in degrees Celsius or Kelvin.

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Caloric content

Equation

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When the heat supplied to liquid or solid ($\Delta Q_s$) is added to a body, we observe a proportional increase of the temperature variation in a liquid or solid ($\Delta T_s$). Therefore, we can introduce a proportionality constant the heat capacity ($C$), known as heat capacity, which establishes the following relationship:

$ \Delta Q = C \Delta T $

$C$
Heat capacity
$J/K$
8482
$\Delta Q_s$
Heat supplied to liquid or solid
$J$
10151
$\Delta T_s$
Temperature variation in a liquid or solid
$K$
10152

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Specific heat

Equation

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The heat capacity is related to microscopic oscillations, so it depends less on mass and more on the number of atoms. For this reason, it makes sense to introduce the concept of the specific heat ($c$), which is calculated as the heat capacity ($C$) per unit of the mass ($M$), as follows:

$ c =\displaystyle\frac{ C }{ M }$

$C$
Heat capacity
$J/K$
8482
$M$
Mass
$kg$
5215
$c$
Specific heat
$J/kg K$
5219

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Caloric content as a function of specific heat

Equation

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The heat supplied to liquid or solid ($\Delta Q_s$) can be calculated with the specific heat ($c$), the mass ($M$) and the temperature variation in a liquid or solid ($\Delta T_s$) using:

$ \Delta Q = M c \Delta T$

$\Delta Q_s$
Heat supplied to liquid or solid
$J$
10151
$M$
Mass
$kg$
5215
$c$
Specific heat
$J/kg K$
5219
$\Delta T_s$
Temperature variation in a liquid or solid
$K$
10152

The heat supplied to liquid or solid ($\Delta Q_s$) is related to the temperature variation in a liquid or solid ($\Delta T_s$) and the heat capacity ($C$) as follows:

$ \Delta Q = C \Delta T $



Where the heat capacity ($C$) can be replaced by the specific heat ($c$) and the mass ($M$) using the following relationship:

$ c =\displaystyle\frac{ C }{ M }$



Therefore, we obtain:

$ \Delta Q = M c \Delta T$

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Specific heat of a system

Equation

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The quantity of the heat capacity ($C$) in a system of the i-th mass of the system ($M_i$) with the specific heat of the i-th mass ($c_i$) can be calculated as follows:

$C = \displaystyle\sum_i c_i M_i$



Therefore, the total sum for the specific heat ($c$) calculated is:

$ c =\displaystyle\frac{ \displaystyle\sum_i c_i M_i }{ \displaystyle\sum_i M_i }$

The quantity of the heat capacity ($C$) in a system of the i-th mass of the system ($M_i$) with the specific heat of the i-th mass ($c_i$) can be calculated as follows:

$C = \displaystyle\sum_i c_i M_i$



where the sum of the masses is obtained as:

$M = \displaystyle\sum_i M_i$



So, with the help of

$ c =\displaystyle\frac{ C }{ M }$

,

we can calculate the heat capacity ($C$) as follows:

$ c =\displaystyle\frac{ \displaystyle\sum_i c_i M_i }{ \displaystyle\sum_i M_i }$

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Soil specific heat

Equation

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The specific heat of the soil depends on variables the dry mass of sand in the sample ($M_a$), the dry mass of silt in the sample ($M_i$), and the dry mass of clay in the sample ($M_c$), in addition to the mass of water in the soil ($M_w$). Together with the specific heat of sand ($c_a$), the specific heat of silt ($c_i$), the specific heat of clay ($c_c$), and the specific heat of water ($c_w$), these variables allow for the calculation of the specific heat of the soil. In particular, we can work with the proportions the mass fraction of sand in the sample ($g_a$), the mass fraction of silt in the sample ($g_i$), the mass fraction of clay in the sample ($g_c$), and the relationship gravimetric water solido ($\theta_w$) and demonstrate that:

$ c = \displaystyle\frac{ g_a c_a + g_i c_i + g_c c_c + \theta_w c_w }{1+ \theta_w }$

With the i-th mass of the system ($M_i$) and the specific heat of the i-th mass ($c_i$), you can calculate the specific heat ($c$) for the soil using the equation:

$ c =\displaystyle\frac{ \displaystyle\sum_i c_i M_i }{ \displaystyle\sum_i M_i }$



Furthermore, using the variables the dry mass of sand in the sample ($M_a$), the dry mass of silt in the sample ($M_i$), the dry mass of clay in the sample ($M_c$), and the mass of water in the soil ($M_w$) along with the specific heat of sand ($c_a$), the specific heat of silt ($c_i$), the specific heat of clay ($c_c$), and the specific heat of water ($c_w$), you can obtain the specific heat (

$c$

) with the formula:

$c=\displaystyle\frac{M_ac_a+M_ic_i+M_cc_c+M_wc_w}{M_a+M_i+M_c+M_w}$



Using the following equations:

$ g_a =\displaystyle\frac{ M_a }{ M_s }$



$ g_i =\displaystyle\frac{ M_i }{ M_s }$



$ g_c =\displaystyle\frac{ M_c }{ M_s }$



$ g_a + g_i + g_c = 1$



and

$ \theta_w =\displaystyle\frac{ M_w }{ M_s }$



Then, the specific heat ($c$) is simplified using the following equation:

$ c = \displaystyle\frac{ g_a c_a + g_i c_i + g_c c_c + \theta_w c_w }{1+ \theta_w }$



The specific heat primarily depends on the water content but also on the texture and, hence, the proportion of sand, silt, and clay in the soil. In any case, the specific heats of the different components are as follows:

Component $c$ [J/kg K]
Sand 830
Silt 1350
Clay 1350
Water 4184

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