Question

How will you measure the height of a building using only a thermometer and a watch?

Answer 1

$$\begin{gathered} \mathrm{In}\mathrm{static}\mathrm{Environment},\mathrm{the}\mathrm{lapse}\mathrm{rate}\mathrm{is}1.{05}^0\mathrm{C}\mathrm{per}100\mathrm{m}. \\ \mathrm{Supposing}\mathrm{you}\mathrm{had}: \\ 1)\mathrm{no}\mathrm{wind}\mathrm{vector}(\mathrm{drafts}) \\ 2)\mathrm{highlty}\mathrm{accurate}\mathrm{thermometer} \\ 3)\mathrm{instantaneous}\mathrm{measurement} \\ 4)\mathrm{insulated}\mathrm{body}\mathrm{suit} \\ \mathrm{Let}\mathrm{the}\mathrm{measured}\mathrm{temperature}\mathrm{at}\mathrm{the}\mathrm{top}\mathrm{of}\mathrm{the}\mathrm{room}:37.{01}^{\mathrm{o}}\mathrm{C} \\ \mathrm{and}\mathrm{the}\mathrm{measured}\mathrm{temperature}\mathrm{at}\mathrm{the}\mathrm{bottom}:36.{97}^{\mathrm{o}}\mathrm{C} \\ \mathrm{Now}, \\ \frac{\left(37.{01}^{\mathrm{o}}-36.{97}^0\right)}{\mathrm{lapse}\mathrm{rate}}=\frac{0.{04}^{\mathrm{o}}}{1.{05}^{\mathrm{o}}}\times 100=3.81\mathrm{m} \end{gathered}$$