For an ideal gas in the Boussinesq limit,
$$
g'=g\frac{\hat\rho}{\rho_0}\simeq g\frac{T-T_0}{T_0}.
$$
Each opening removes $\rho_0c_pV_{\rm ex}(T-T_0)$ of heat. With $n$ openings per hour, the mean removal rate is $(n/3600)$ times this quantity. Equating it to $\dot q$ and writing $\Delta T=T-T_0$ gives
$$
\dot q=\frac n{3600}\frac{\rho_0c_pWH\delta t}{4}
\sqrt{\frac{gH}{T_0}}\,(\Delta T)^{3/2}.
$$
Therefore
$$
\boxed{
T=T_0+
\left[
\frac{14400\,\dot q}{n\rho_0c_pWH\delta t}
\sqrt{\frac{T_0}{gH}}
\right]^{2/3}}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 345 / 3 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-345/3/e/solution