Respuesta :
Answer:
38 canisters
Explanation:
The combustion of propane can be represented by the equation below:
C₃H₈ (g) + O₂ (g) → CO₂ (g) + H₂O(g) ΔHc = -103.85 kJ/mol
The boiling of water is represented below:
H₂O(l) → H₂O(g) ΔH = Q = mcΔT
The heat necessary to make water reach the boiling point is Q, which is calculated with the mass of water (m = 4.0 kg), its specif heat capacity ( c = 4.200 J/kg°C) and temperature variation (ΔT = 100 - 25)
Therefore, ΔH = Q = 4.0 kg x 4.200 kJ/kg°C x 75 °C
ΔH = Q = 1260.0 kJ
This is the amount of heat necessary to heat the water every day. Since the expedition will last 7 days, the total heat will be 8820 kJ
Now we need to calculate the amount of propane needed to generate this heat:
1 mol C₃H₈ _______ 103.85 kJ
X _______ 8820 kJ
x = 84.9 mol C₃H₈
1 mol C₃H₈ _____ 44 g
84.9 mol C₃H₈ ___ x
x = 3735.6 g
1 canister _____ 100 g C₃H₈
x _____ 3735.6 g
x = 37.3 canisters
The minimum number of fuel canisters I must bring is 38 canisters.
37.35 canisters will be required to boil 4kg water for 7 days.
The combustion reaction of propane,
[tex]\rm \bold{ C_3H_8 (g) + O_2 (g) \rightarrow CO_2 (g) + H_2O(g) }[/tex]
The energy require to boil the water
[tex]\rm \bold{ Q = mc\Delta T}[/tex]
Where,
Q is energy required
m- mass =4kg
c - specific heat capasity = [tex]\rm \bold { 4.200 J/kg^\cdot C}[/tex]
hence the enrgy required for 7 days will be 8820 kJ
As we know,
The molar mass of propane = 44g/mol
1mol of propane generate 103.85 kJ of heat.
1 canister = 100 g
To genrate 8820 kJ of energy 84.9 moles of propane will needed.
Hence we need 3735.6 g of propane.
The number of canisters we need,
[tex]\rm \bold{canisters = \frac{3735.6 }{100} }\\\\\rm \bold{canisters = 37.35}[/tex]
So we can say that we need 37.35 canisters to boil 4kg water for 7 days.
To know more about Specific heat capacity, refer to the link:
https://brainly.com/question/11194034?referrer=searchResults