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Let's say the temperature of an object is constant and equal to 20 °C. If we put this object to a surrounding which has a temperature of 25 °C, than there's heat transfer from the surrounding to object by convection. If $q$ is the heat transfer (in Watt) than after a specific time ($Δ$t) we will have Temperature of the object equal to 25 °C ( If we consider the temperature of the surrounding constant) and thus we will have a net heat transfer equal to $Q = qxΔt$ if $q$ is constant. My question is, if we consider a 1D heat transfer, the surface of the object exposed to the surrounding will get the heat and thus will receive each $dt$ the amount of heat $q$ ($q$ is measured like a current, in a cross section of the object) then will gain as internal energy $\Delta U = q$ but we say that all the object gained the heat $q$ not only the surface exposed to the surrounding. Can we say, first the surface of the object exposed will gain an internal energy equal to $q$ (increasing its temperature) and thus "distributing" its energy "uniformly" to the other parts of the object so we can say that all the object gained $\Delta U$ as energy ?

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Let's say the temperature of an object is constant and equal to 20 °C. If we put this object to a surrounding which has a temperature of 25 °C, than there's heat transfer from the surrounding to object by convection. If $q$ is the heat transfer (in Watt) than after a specific time ($Δt$) we will have Temperature of the object equal to 25 °C.

This is not entirely correct.

Take an object of mass $m$ and heat capacity $c_p$ at an initial temperature $T_0$. At $t=0$ we place it in surroundings of constant $T_s (> T_0)$.

Heat now starts to flow from the surroundings to the object. In an infinitesimal amunt of time $dt$ with Newton's heating/cooling Law:

$dq=kA(T_s-T(t))dt$, where $k$ is the heat transfer coefficient, $A$ the surface area of the object and $T(t)$ the instantaneous temperature of the object.

This increment in Enthalpy $dq$ raises the temperature of the object by $dT$ acc.:

$dq=mc_pdT(t)$.

So we have:

$mc_pdT(t)=kA(T_s-T(t))dt$, a differential equation where variables can be separated.

Integrated between $t=0, T_0$ and $t, T(t)$ we get:

$T(t)=T_s-(T_s-T_0)e^{-\alpha t}$, with $\alpha=\frac{kA}{mc_p}$.

What this shows is that $T(t)=T_s$ only for $t \to +\infty$ because only then does $e^{-\alpha t} \to 0$.

So in reality $\Delta t= +\infty$! This is a very frequent outcome in this kind of heating/cooling situations.

As regards:

Can we say, first the surface of the object exposed will gain an internal energy equal to q (increasing its temperature) and thus "distributing" its energy "uniformly" to the other parts of the object so we can say that all the object gained $ΔU$ as energy?

The outer surface layer of the object first heats up but the Enthalpy it absorbed then flows from that hotter surface to deeper, colder parts by means of heat conduction. The object thus has a heat gradient ($\frac{dT}{dx}$, in the $x$ direction for example) from its boundary to its core. Over time that gradient flattens, until the object has a homogeneous temperature.

It's not really necessary to use Internal energy here: as no work is being performed: $\Delta U = q$.

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