Question

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

Answer 1

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.