Question

At a festival, spherical balloons with a radius of 280. Cm are to be inflated with hot air and released. The air at the festival will have a temperature of 25° C and must be heated to 100°C to make the balloons float. 3.00 kg of propane (C3H8) fuel are available to burn to heat the air. Calculate the number of balloons that can be inflated with hot air.

Answer 1

Answer:

8608.18 balloons

Explanation:

Hello! Let's solve this!

Data needed:

Enthalpy of propane formation: 103.85kJ / mol

Specific heat capacity of air: 1.009J · g ° C

Density of air at 100 ° C: 0.946kg / m3

Density of propane at 100 ° C: 1.440kg / m3

First we will calculate the propane heat (C3H8)

3000g * (1mol / 44g) * (103.85kJ / mol) * (1000J / 1kJ) = 7.08068 * 10 ^ 6 J

Then we can calculate the mass of the air with the heat formula

Q = mc delta T

m = Q / c delta T = (7.08068 * 10 ^ 6 J) / (1.009J / kg ° C * (100-25) ° C) =

m = 93566.96kg

We now calculate the volume of a balloon.

V = 4/3 * pi * r ^ 3 = 4/3 * 3.14 * 1.4m ^ 3 = 11.49m ^ 3

Now we calculate the mass of the balloon

mg = 0.946kg / m3 * 11.49m ^ 3 = 10.87kg

The amount of balloons is

93566.96kg / 10.87kg = 8608.18 balloons

Answer 2

The question is incomplete, here is the complete question:

At a festival, spherical balloons with a radius of 280 cm are to be inflated with hot air and released. The air at the festival will have a temperature of 25° C and must be heated to 100°C to make the balloons float. 3.00 kg of propane (C₃H₈) fuel are available to burn to heat the air. Calculate the number of balloons that can be inflated with hot air.

Density of air at 100°C = $0.946kg/m^3$

Specific heat capacity of air = 1.009 J/g°C

Answer: The number of balloons inflated with hot air are 7.

Explanation:

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

The chemical equation for the combustion of propane follows:

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(3\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(C_3H_8(g))})+(5\times \Delta H_f_{(O_2(g))})]$

We are given:

$\Delta H_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.51kJ/mol\\\Delta H_f_{(C_3H_8(g))}=-103.8kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(3\times (-393.51))+(4\times (-241.8))]-[(1\times (-103.8))+(5\times (0))]\\\\\Delta H_{rxn}=-2043.93kJ/mol$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of propane = 3.00 kg = 3000 g    (Conversion factor:  1 kg = 1000 g)

Molar mass of propane = 44 g/mol

Putting values in above equation, we get:

$\text{Moles of propane}=\frac{3000g}{44g/mol}=68.18mol$

To calculate the heat released, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat released = ? kJ

n = number of moles = 68.18 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = -2043.93 kJ/mol

Putting values in above equation, we get:

$-2043.93kJ/mol=\frac{q}{68.18mol}\\\\q=(-2043.93kJ/mol\times 68.18mol)=-1.39\times 10^5kJ$

To calculate the heat released by the reaction, we use the equation:

$q=mc\Delta T$

where,

q = heat released = $1.39\times 10^5kJ=1.39\times 10^8J$

m = mass of air = ?

c = heat capacity of air = 1.009 J/g°C

$\Delta T$ = change in temperature = $T_2-T_1=(100-25)^oC=75^oC$

Putting values in above equation, we get:

$1.39\times 10^8J=m\times 1.009J/g^oC\times 75^oC\\\\m=\frac{1.39\times 10^8}{1.009\times 75}=1.84\times 10^6g=1.84\times 10^3kg$

Conversion factor: 1 kg = 1000 g

To calculate the volume of air, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of air = $0.946kg/m^3$

Mass of air = $1.84\times 10^3kg$

Putting values in above equation, we get:

$0.946kg/m^3=\frac{1.84\times 10^3kg}{\text{Volume of air}}\\\\\text{Volume of air}=\frac{1.84\times 10^3kg}{0.946kg/m^3}=1945.03m^3$

To calculate the volume of sphere, we use the equation:

$V=4\pi r^3$

where,

r = radius of the sphere = 280 cm = 2.8 m    (Conversion factor: 1 m = 100 cm)

Putting values in above equation, we get:

$V=4\times 3.14\times (2.8)^3\\\\V=275.7m^3$

To calculate the number of balloons inflated, we divide the volume of air by the volume of 1 balloon:

$\text{Number of balloons inflated}=\frac{V_{air}}{V_{balloon}}\\\\\text{Number of balloons inflated}=\frac{1945.03m^3}{275.7m^3}=7$

Hence, the number of balloons inflated with hot air are 7.