LP gas burns according to the exothermic reaction: C3H8(g) + 5 O2(g) right arrow 3 CO2(g)...

LP gas burns according to the exothermic reaction:
C3H8(g) + 5 O2(g) right arrow 3 CO2(g) + 4 H2O(g) ΔHrxn° = −2044 kJ
What mass of LP gas is necessary to heat 1.8 L of water from room temperature (25.0°C) to boiling (100.0°C)? Assume that during heating, 16% of the heat emitted by the LP gas combustion goes to heat the water. The rest is lost as heat to the surroundings. (See the table.)

Solutions

Expert Solution

1.8 L water = 1800 ml = 1800 gm(density of water is 1)

specific heat of water = 4.186 J/gm

tempreture diffrence = 100 - 25 = 75⁰ C

q = mass specific heat tempreture diffrence

substitute value in above equation

q = 1800 4.186 75

q = 565110 J = 565.11 KJ

565.11 KJ heat require to raise tempreture of 1.8 L water from 25⁰C to 100⁰C

16% heat emmited by LP gas goes to heat water thus, total heat should produce by LP gas = 100565.11/16 = 3531.9375 KJ

1 mole of C₃H₈ produce 2044 KJ heat then to produce 3531.9375 KJ heat require = 3531.9375/2044 = 1.7279 mole of LP gas

molar mass of C₃H₈ = 44.1g/mole that meaan 1 mole of C₃H₈ = 44.1 gm then 1.7279 mole = 1.727944.1 = 76.20gm

76.20gm of LP gas is necessary to heat 1.8 L of water from room temperature (25.0°C) to boiling (100.0°C)

queen_honey_blossom answered 1 year ago

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