LP gas burns according to the exothermic reaction:C3H8(g)+5 O2(g)→3 CO2(g)+4 H2O(g)ΔH∘rxn=−2044 kJ 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, 15% of the heat emitted by the LP gas combustion goes to heat the water. The rest is lost as heat to the surroundings.

Question
Asked Oct 26, 2019

LP gas burns according to the exothermic reaction:

C3H8(g)+5 O2(g)→3 CO2(g)+4 H2O(g)ΔH∘rxn=−2044 kJ

 

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, 15% of the heat emitted by the LP gas combustion goes to heat the water. The rest is lost as heat to the surroundings.

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Expert Answer

Step 1

Given:

The balanced chemical equation:

C3H8(g)+5 O2(g)→3 CO2(g)+4 H2O(g)

ΔHrxn=−2044 kJ

Volume of water = 1.5 L

Heat emitted = 15%

Change in T = (25.0 °C) to (100.0 °C)

Step 2

1.5L means 1500 ml since density of water = 1,

1 ml = 1 g

1500 ml = 1500 gr

The heat absorbed by water (Q) = 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...

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