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Thermal Balance
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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.
ID:(312, 0)
Second Law of Thermodynamics
Description
$dS=\displaystyle\displaystyle\frac{\delta Q}{T}$
ID:(3203, 0)
Thermal Balance
Description
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.
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Equations
(ID 3197)
Since the mechanical force ($F$) divided by the section ($S$) is equal to the pressure ($p$):
| $ p \equiv\displaystyle\frac{ F }{ S }$ |
and the volume Variation ($\Delta V$) with the distance traveled ($dx$) is equal to:
| $ \Delta V = S \Delta s $ |
The equation for the differential inexact labour ($\delta W$) can be expressed as:
| $ \Delta W = F \Delta s $ |
So it can be written as:
| $ \delta W = p dV $ |
(ID 3468)
Since the internal energy differential ($dU$) relates to the differential inexact Heat ($\delta Q$) and the differential inexact labour ($\delta W$) as shown below:
| $ dU = \delta Q - \delta W $ |
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:
| $ \delta W = p dV $ |
Therefore, we can conclude that:
| $ dU = \delta Q - p dV $ |
(ID 3470)
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:
| $ W =\displaystyle\int_C \vec{F} \cdot d \vec{s} $ |
(ID 3601)
Examples
(ID 678)
$\displaystyle\frac{dU}{dt}= MET m$
(ID 3616)
(ID 679)
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:
| $ \Delta Q = C \Delta T $ |
(ID 3197)
$\eta=\displaystyle\frac{\delta W}{dU}$
(ID 3617)
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:
| $ dW = \vec{F} \cdot d\vec{s} $ |
In the continuous limit, the sum can be represented as an integral:
| $ W =\displaystyle\int_C \vec{F} \cdot d \vec{s} $ |
(ID 3601)
(ID 677)
$\delta Q=(1-\eta)dU$
(ID 3618)
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):
| $ dU = \delta Q - \delta W $ |
(ID 9632)
The differential inexact labour ($\delta W$) is equal to the pressure ($p$) multiplied by the volume Variation ($\Delta V$):
| $ \delta W = p dV $ |
(ID 3468)
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:
| $ dU = \delta Q - p dV $ |
(ID 3470)
$\delta Q_c=\delta Q_t+\delta Q_r+\delta Q_e$
(ID 3621)
$dS=\displaystyle\displaystyle\frac{\delta Q}{T}$
(ID 3203)
ID:(312, 0)