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Sagot :
Answer:
272° C
Explanation:
Given :
Volume of the balloon, V = 500 [tex]m^3[/tex]
The temperature of the surrounding air, [tex]T_{air} = 15^\circ C[/tex]
Total load, [tex]m_{T}[/tex] = 290 kg
Density of the air, [tex]$\rho_{air} = 1.23 \ kg/m^3$[/tex]
We known buoyant force,
[tex]$F_B = \rho_{air} V$[/tex]
For a 290 kg lift, [tex]$m_{hot} = \frac{F_B}{g} = 290 \ kg$[/tex]
[tex]$m=\rho V$[/tex]
∴ [tex]$m_{hot}=\rho_{hot} V ; \ \ \ \ \ \frac{F_B}{g}-m_{hot} = 290 \ kg$[/tex]
[tex]$(\rho_{air} - \rho_{hot}) V= 290 \ kg$[/tex]
[tex]$\rho_{hot} = \rho_{air}- \frac{290}{V} \ kg = 1.23 \ kg/m^3 - \frac{290 \ kg}{500 \cm^3}$[/tex]
[tex]$\rho_{hot}= 0.65 \ kg/m^3 =\frac{\rho M}{R T_{hot}}$[/tex]
∴ [tex]$\rho_{hot} T_{hot}= \rho_{air} T_{air}$[/tex]
[tex]$T_{hot}= T_{air}\left[\frac{\rho_{air}}{\rho_{hot}}\right]$[/tex]
[tex]$=288 \ K \times \frac{1.23 \ kg/m^3}{0.65 \ kg/m^3}$[/tex]
= 545 K
[tex]$=272^\circ C$[/tex]
Therefore, temperature of the air in the balloon is 272 degree Celsius.
To lift a load more than the weight of the balloon, the temperature of the air in the balloon has to be higher than the air in the surrounding.
- The temperature of the air in the balloon to lift a total load of 290 kg is approximately 272.12°C.
Reasons:
Given information are;
Volume of the balloon = 500.0 m³
Temperature of the surrounding air = 15.0°C
Density of air at 15.0°C = 1.23 kg/m³
Required:
The temperature required to lift 290kg.
Solution:
Let, [tex]\rho _{air , b}[/tex], represent the density of the air in the balloon, we have;
[tex]\rho _{air , b}[/tex] × 500.0 + 290 = 1.23 × 500
Therefore;
[tex]\displaystyle \rho _{air , b} = \frac{1.23 \times 500- 290}{500} = 0.65[/tex]
According to the Ideal Gas Law, we have;
ρ₁ × R × T₁ = ρ₂ × R × T₂
Therefore;
[tex]\displaystyle T_2 = \mathbf{\frac{\rho_1 \times T_1}{\rho_2}}[/tex]
Therefore;
[tex]\displaystyle T_2 = \frac{1.23\times288.15}{0.65} \approx 545.27[/tex]
The temperature of the balloon, T₂ ≈ 545.27 - 273.15 = 272.12
The temperature of the air in the balloon, T₂ ≈ 272.12 °C
Learn more hear:
https://brainly.com/question/11236279
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