Thermal Balance

Storyboard

Eating allows us to generate energy to perform different activities. However, the process is not efficient, losing a part of the heat energy that the body must eliminate. Elimination occurs by three possible paths; transport, radiation and evaporation.

>Model

ID:(312, 0)



Obtained Energy

Definition

ID:(678, 0)



MET Values

Image

ID:(679, 0)



Second Law of Thermodynamics

Quote

$dS=\displaystyle\displaystyle\frac{\delta Q}{T}$

ID:(3203, 0)



Thermal Balance

Storyboard

Eating allows us to generate energy to perform different activities. However, the process is not efficient, losing a part of the heat energy that the body must eliminate. Elimination occurs by three possible paths; transport, radiation and evaporation.

Variables

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$\delta Q_t$
dQ_t
Body Heat Eliminated Transportation
J
$\delta Q_r$
dQ_r
Body Heat Removed by Radiation
J
$\delta Q_e$
dQ_e
Body Heat Removed by Sweat
J
$\delta Q$
dQ
Differential inexact Heat
J
$\delta W$
dW
Differential inexact labour
J
$\eta$
eta
Efficiency
-
$\vec{F}$
&F
Force
N
$C$
C
Heat capacity
J/K
$\Delta Q$
DQ
Heat supplied to liquid or solid
J
$\delta Q_c$
dQ_c
Heat to be Removed from the Body
J
$dt$
dt
Infinitesimal Variation of Time
s
$M$
M
Mass
kg
$MET$
MET
Metabolic Equivalent
$\vec{s}$
&s
Posición (vector)
m
$p$
p
Pressure
Pa
$\Delta T$
DT
Temperature variation
K
$\delta Q$
dQ
Variation of heat
J
$dU$
dU
Variation of the internal energy
J
$\Delta V$
DV
Volume Variation
m^3
$W$
W
Work
J

Calculations


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

Symbol
Equation
Solved
Translated

Calculations

Symbol
Equation
Solved
Translated

 Variable   Given   Calculate   Target :   Equation   To be used



Equations

Since the mechanical force ($F$) divided by the section ($S$) is equal to the pressure ($p$):

equation=4342

and the volume Variation ($\Delta V$) with the distance traveled ($dx$) is equal to:

equation=4346

The equation for the differential inexact labour ($\delta W$) can be expressed as:

equation=3202

So it can be written as:

equation

Since the internal energy differential ($dU$) relates to the differential inexact Heat ($\delta Q$) and the differential inexact labour ($\delta W$) as shown below:

equation=9632

And it is known that the differential inexact labour ($\delta W$) is related to the pressure ($p$) and the volume Variation ($\Delta V$) as follows:

equation=3468

Therefore, we can conclude that:

equation

Expanding the concept to a longer path involves summing up the energy required for each path element:

$\bar{W}=\displaystyle\sum_i \vec{F}_i\cdot\Delta\vec{s}_i$



However, the value of this equation represents just an average energy required or generated. The precise energy is obtained when the steps become very small, allowing the force to be considered constant within them:

$W=\displaystyle\sum_i \mbox{lim}_{\Delta\vec{s}_i\rightarrow\vec{0}}\vec{F}_i\cdot\Delta\vec{s}_i$



In this limit, the energy corresponds to the integral along the traveled path, giving us:

equation


Examples

$\displaystyle\frac{dU}{dt}= MET m$

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

kyon

$\eta=\displaystyle\frac{\delta W}{dU}$

Carnot was the first to describe energy in terms of the path and the necessary force to traverse it. Progressing along a path with a force requires or generates energy. This corresponds to the equation:

equation=1136

In the continuous limit, the sum can be represented as an integral:

kyon

$\delta Q=(1-\eta)dU$

The internal energy differential ($dU$) is always equal to the amount of the differential inexact Heat ($\delta Q$) supplied to the system (positive) minus the amount of the differential inexact labour ($\delta W$) performed by the system (negative):

kyon

The differential inexact labour ($\delta W$) is equal to the pressure ($p$) multiplied by the volume Variation ($\Delta V$):

kyon

With the first law of thermodynamics, it can be expressed in terms of the internal energy differential ($dU$), the differential inexact Heat ($\delta Q$), the pressure ($p$), and the volume Variation ($\Delta V$) as:

kyon

$\delta Q_c=\delta Q_t+\delta Q_r+\delta Q_e$

$dS=\displaystyle\displaystyle\frac{\delta Q}{T}$


>Model

ID:(312, 0)