LP gas burns according to the following exothermic reaction: C3H8(g)+5O2(g)→3CO2(g)+4H2O(g)ΔH∘rxn=−2044kJ What mass of LP gas is necessary to heat 1.5 L of water from room temperature (25.0 ∘C) to boiling (100.0 ∘C)? Assume that during heating, 14% of the heat emitted by the LP gas combustion goes to heat the water. The rest is lost as heat to the surroundings
find ot the fist mass of water
1.5L means 1500 ml since density of water = 1, 1 ml = 1 graso 1500 ml = 1500 gr
heat abosrbed by water = m x c x delta T
m = mass of the water = 1500 gr
c - specific heat of the water = 4.184 J/g ºC
delta T = T2-T1 = 100-75
put all these in the above equation
heat abosrbed by water = 1500 g x 4.184 J/g ºC x (100 ºC-25 ºC) = 471,000 J = 471 kJ
471 kJ x (1 mole C3H8 / 2044 kJ) x (44.0 g C3H8 / 1 mole C3H8) = 10.1 g C3H8
Since it is only 14% efficient,
10.1 g C3H8 x (100 / 14) = 72.143 g C3H8.
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