To compute the temperature of the air inside the hot air balloon, we can use the ideal gas law. The principles of buoyancy and the ideal gas law give us a relationship between the pressures and temperatures of the air inside and outside the balloon.
The key relationship we can use is based on the following equation derived from the ideal gas law:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
Where:
- \(P_1\) is the pressure inside the balloon (assuming it is approximately equal to the external pressure plus the buoyancy based on lifting capacity)
- \(T_1\) is the temperature of the air inside the balloon (in Kelvin)
- \(P_2\) is the external pressure (sea level pressure)
- \(T_2\) is the external temperature (in Kelvin)
Firstly, we need to convert the temperatures from Celsius to Kelvin. The conversion is done using:
\[ T(K) = T(°C) + 273.15 \]
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Convert outside temperature to Kelvin:
\[ T_2 = 10.0 + 273.15 = 283.15 , K \]
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Convert pressure from hPa to Pa:
Since 1 hPa = 100 Pa, the sea level pressure is:
\[ P_2 = 1013.25 , hPa \times 100 = 101325 , Pa \]
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Estimate the pressure inside the balloon \(P_1\):
We know that the lifting capacity includes the weight of the balloon, the basket, and any payload.
The total weight supported by the balloon is given by:
\[ W = \text{mass} \times g \]
where \( g \) (acceleration due to gravity) is approximately \( 9.81 , m/s^2 \).
Given that the total mass (balloon, basket, and payload) is 780 kg, we have:
\[ W = 780 , kg \times 9.81 , m/s^2 = 6093.8 , N \]
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Calculating the volume of the balloon:
The volume \( V \) of a cylinder (approximately) can be calculated using the formula for the volume of a cylinder:
\[ V = \pi r^2 h \]
where \( r \) is the radius and \( h \) is the height. The width is given as 14 m, making the radius \( r = 7 , m \) and the height \( h = 23 , m \).
\[ V = \pi (7)^2 (23) \approx \pi (49)(23) \approx 3555.86 , m^3 \]
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Using the buoyancy equation:
The buoyant force can be calculated as:
\[ \text{Buoyant Force} = V \cdot \Delta \rho \cdot g \]
where \( \Delta \rho \) is the density difference between the outside air and the hot air inside the balloon. We want to set the buoyancy equal to the weight:
\[ V \cdot \Delta \rho \cdot g = 6093.8 , N \]
To find \( \Delta \rho \):
- The density of air can be computed as:
\[ \rho = \frac{P}{R \cdot T} \]
where \( R \) is the specific gas constant for dry air, approximately \( 287 , J/(kg \cdot K) \).
For the outside air density:
\[ \rho_2 = \frac{101325 , Pa}{287 , J/(kg \cdot K) \cdot 283.15 , K} \approx 1.225 , kg/m^3 \]
Let's denote the inner density as \( \rho_1 \):
We know that:
\[ \Delta \rho = \rho_2 - \rho_1 \]
We substitute:
\[ 3555.86 m^3 \cdot (\rho_2 - \rho_1) \cdot 9.81 = 6093.8 \]
Rearranging gives:
\[ \rho_2 - \rho_1 = \frac{6093.8}{3555.86 \cdot 9.81} \approx 0.168 , kg/m^3 \]
Thus:
\[ \rho_1 = \rho_2 - 0.168 \approx 1.225 - 0.168 \approx 1.057 , kg/m^3 \]
Finally, using the inner density formula to find \( T_1 \):
\[ \rho_1 = \frac{P_1}{R \cdot T_1} \]
Rearranging gives:
\[ T_1 = \frac{P_1}{R \cdot \rho_1} \]
Now we need \( P_1\) approximately equal the external pressure \( P_2 \) because the balloon is not fully pressurized. Assuming \( P_1 \approx P_2 = 101325 \), we get:
\[ T_1 = \frac{101325}{287 \cdot 1.057} \approx 344.8 , K \]
Converting back to Celsius:
\[ T_1(°C) = T_1(K) - 273.15 = 344.8 - 273.15 \approx 71.65 °C \]
Final Result:
The temperature of the air inside the balloon is approximately 71.65 °C.