Answer:
72.25 g
Explanation:
mass of 1.7 L water = 1.7 x 10⁻³ x 10³ kg
= 1.7 kg
heat required to raise its temperature from 25 degree to 100 degree
= mass x specific heat x rise in temperature
= 1.7 x 4.18 x 10³ x 75 J
= 532.95 kJ
Now calorific value of LP gas = 46.1 x 10⁶J / kg
Let required mass of LP gas be m kg
heat evolved from this amount of gas
= 46.1 x 10⁶ m
Heat utilized in warming water
= 46.1 x 10⁶ m x .16 J
So
46.1 x 10⁶ m x .16 = 532.95 x 10³
7376 x 10³ m = 532.95 x 10³
m = 532.95 / 7376 kg
= 72.25 g
The mass of LP required is 0.0018 g of LP
C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)
ΔH∘rxn= −2044kJ
We know from the question that the volume of water = 1.7 L
Since the density of water is 1 g/L
Mass of water = 1.7 g
Heat capacity of water = 4.18 J/g.∘C
Heat gained by water = mcθ
m = mass of water
c = heat capacity of water
θ = temperature rise of water
H = 1.7 g × 4.18 J/g.∘C × (100.0 ∘C - 25.0 ∘C)
H = 532.95 J
Heat lost by LP = heat gained by water
Recall that only 16% of the heat lost by LP was gained by water so, heat lost by LP = 0.16 × (- 532.95 J) = -85.827 J
1 mole of LP contains 44 g
44 g of LP produces −2044 × 10^3 J of heat
x g of LP produces -85.827 J
x = 44 g × -85.827 J /−2044 × 10^3 J
x = 0.0018 g of LP
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