A hot air balloon with a payload rises in the | vedclass

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(C) For the hot air balloon to rise,the buoyant force (upthrust) must be greater than or equal to the total weight of the balloon,payload,and the hot

Vedclass · Accepted Answer

$105$

Vedclass · Answer

(C) For the hot air balloon to rise,the buoyant force (upthrust) must be greater than or equal to the total weight of the balloon,payload,and the hot air inside.Let $V$ be the volume of the balloon,$ho_o$ be the density of outside air,and $ho_i$ be the density of inside hot air.$V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (5.85)^3 \approx 838.5 \, m^3$.Using the ideal gas law,$PV = nRT = \frac{m}{M} RT$,so $ho = \frac{m}{V} = \frac{PM}{RT}$.Equating forces: $V ho_o g = V ho_i g + m_{payload} g$.$V(ho_o - ho_i) = 210$.$V \frac{PM}{R} \left( \frac{1}{T_o} - \frac{1}{T_i} ight) = 210$.Substituting values: $838.5 	imes \frac{10^5 	imes 30 	imes 10^{-3}}{8.31} \left( \frac{1}{300} - \frac{1}{T_i} ight) = 210$.$302647 \left( \frac{1}{300} - \frac{1}{T_i} ight) = 210$.$\frac{1}{300} - \frac{1}{T_i} \approx 0.0006938$.$\frac{1}{T_i} \approx 0.003333 - 0.0006938 = 0.002639$.$T_i \approx 378.9 \, K$.Converting to Celsius: $T_i \approx 378.9 - 273 = 105.9^{\circ} C$.Thus,the temperature is close to $105^{\circ} C$.

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