Take a system picture as below.
- $\dot{V}$ - air flow (m$^3$/s)
- $\rho$ - air density (kg/m$^3$)
- $\tilde{C}_p$ - air specific heat (J/kg $^o$C)
- $h_a$ - air convection coefficient (W/m$^2$ $^o$C)
- $T_a$ - air temperature (K)
- $A$ - area of container (m$^2$)
- $w$ - wall thickness (m)
- $k$ - wall thermal conductivity (W/m K)
- $\hat{\dot{q}}_I$ - internal heat generation (W/m$^3$)
- $V$ - volume of container (m$^3$)
- $T_f$ - unknown final equilibrium temperature
Assume that heat generation is uniform throughout the container, the container is (mostly) air, and the air is well mixed. The energy balance equation gives
$$\hat{\dot{q}}_I V = \left(\dot{V}\rho\tilde{C}_p + (A/R_T)\right)\left(T_f - T_a\right)$$
where $R_T$ is a thermal resistance defined as
$$R_T = \left(\frac{w}{k} + \frac{1}{h_a}\right)$$
Cast this into the following form
$$\left(\frac{T_f}{T_a} - 1 \right) = \left[\alpha\ \hat{\dot{q}}_I\right] \left[1 + \alpha \left(\frac{A}{V}\right)\left(\frac{1}{R_T}\right) \right]^{-1}$$
with
$$\alpha = \frac{\tau}{\rho \tilde{C}_p T_a}$$
$$ \tau = \frac{V}{\dot{V}}$$
Consider the left side as a relative difference in the temperature inside the container versus the air (this can be expressed as a percentage when multiplied by 100). Consider $\tau$ as a turnover time (how often does one entire volume of air in the container get exchanged).
Now take two extremes.
Case A
In the limit that you do not ventilate, $\tau \rightarrow \infty$. For large $\alpha$, the expression approximates as
$$\left(\frac{T_f}{T_a} - 1 \right) \approx \left[\frac{A}{V}\right]^{-1}\left[R_T\right] \left[\hat{\dot{q}}_I\right]$$
To keep the inside of the container closest to the external air temperature, have the greatest wall area to container volume and have a low thermal resistance. As an extreme analogy, using a thin walled metal sphere that houses all the chickens would be the best option.
Case B
In the limit that you ventilate well, $\tau \rightarrow 0$. For small $\alpha$, the expression approximates as
$$\left(\frac{T_f}{T_a} - 1 \right) \approx \left[\tau\right]\left[\rho\tilde{C}_p T_a\right]^{-1} \left[\hat{\dot{q}}_I\right]$$
To keep the inside of the container closest the external air temperature, have a high air flow (low $\tau$) and replace the air with a denser gas that has a higher heat capacity. So again by extreme analogy, do not ventilate the container with helium (for other reasons as well I can imagine :-)).
Summary
- Use a container with the greatest area to volume possible
- Use a container that has low thermal resistance (alternatively blow the air around the outside of the container as well to increase $h_a$)
- Ventilate well
Interestingly, Case B suggests that, all else being equal, cooling will be more efficient on cooler days than on hot days.
