Thermal exchange between warm object and ideal gas

[JayBi]

Homework Statement

A warm object of mass m and thermal and specific heat c, with temperature T2, is inserted in an adiabatic chamber filled with 1 mol of an ideal (monoatomic) gas at a temperature T1 and atmospheric pressure.
Let T2 > T1.
The chamber has a massless piston, moving without friction.
Considering neglectable the volume of the object, find the equilibrium temperature.

Relevant Equations

1st principle of thermodynamics
Equation of state of an ideal gas

The total variation of internal energy states
$\Delta U_{tot} = 0$
$\Delta U_{tot} = \Delta U_{ext} + \Delta U_{gas} + \Delta U_{obj}$
with:
$\Delta U_{ext} = Q_{ext} - W_{ext} = -W_{ext}$
$\Delta U_{gas} = Q_{gas} - W_{gas} = Q_{gas} - ( - W_{ext} ) = Q_{gas} + W_{ext}$
$\Delta U_{obj} = Q_{obj} - W_{obj} = Q_{obj}$
hence
$0 = ( - W_{ext} ) + ( Q_{gas} + W_{ext} ) + ( Q_{obj})$
$0 = Q_{gas} + Q_{obj}$
Being
$\Delta U_gas = n c_v (T_eq - 1) = Q_gas + W_ext = Q_gas - p_atm (V_eq - V_1) = Q_gas - n R (T_eq - T1)$
We get
$Q_{gas} = n (cv + R) (T_{eq} - T_1) = n cp (T_{eq} - T_1)$
And being
$Q_{obj} = c m (T_eq - T_2)$
We finally get
$T_{eq} = \frac{n cp T_1 + c m T_2}{n cp + c m}$

My question is: the conservation of the internal energy applies as I showed or should I use it for the object and the gas only:
$\Delta U_{gas} + \Delta U_{ext} = 0$
$n cv (T_{eq} - T_1) + c m (T_eq - T_2) = 0$
Arriving at:
$T_{eq} = \frac{n cv T_1 + c m T_2}{n cv + c m}$
With similar steps?
Thanks

 

[vela]

Just follow where the energy goes: the heat released by the object can go to either heating the gas or allowing the gas to do work on the surroundings. Mathematically, we have
\begin{align*}-Q_{\rm obj}&=Q_{\rm gas}\\
-mc(T_{\rm eq}-T_2) &= \underbrace{nC_V(T_{\rm eq}-T_1)}_{\Delta U_{\rm gas}} + \underbrace{p (V_{\rm eq} - V_1)}_{W_{\rm gas}} \\
&= nC_V(T_{\rm eq}-T_1) + n R (T_{\rm eq} - T_1)\\
&= nC_p(T_{\rm eq}-T_1),
\end{align*} which is exactly what you came up with in your first approach.

 

[JayBi]

Just follow where the energy goes: the heat released by the object can go to either heating the gas or allowing the gas to do work on the surroundings. Mathematically, we have
\begin{align*}-Q_{\rm obj}&=Q_{\rm gas}\\
-mc(T_{\rm eq}-T_2) &= \underbrace{nC_V(T_{\rm eq}-T_1)}_{\Delta U_{\rm gas}} + \underbrace{p (V_{\rm eq} - V_1)}_{W_{\rm gas}} \\
&= nC_V(T_{\rm eq}-T_1) + n R (T_{\rm eq} - T_1)\\
&= nC_p(T_{\rm eq}-T_1),
\end{align*} which is exactly what you came up with in your first approach.

Thank you Vela!
My doubt started following the first principle, that links internal energy, not heat directly.
Considering the whole system (gas + body + external) seemed the right approach.
Thank you