If all the energy obtained from burning 0.840 pounds of propane is used to heat water,...

If all the energy obtained from burning 0.840 pounds of propane is used to heat water, how many kilograms of water can be heated from 22.0 °C to 46.8 °C?

Solutions

Expert Solution

Balanced chemical reaction of burning of propane and energy evolved are as given below

C₃H₈ + 5O₂ 3CO₂ + 4H₂O H⁰_rxn = -2218.6 KJ

molar mass of propane = 44.1 gm/mole

0.840 pound of propane = 381.0176 gm

mole of propane in 381.0176 gm = 381.0176 / 44.1 = 8.64 mole

According to reaction 1 mole propane produce heat = -2218.6 KJ then 8.64 mole porpane produce heat =

-2218.6 X 8.64 = -19168.4 KJ

heat released by 0.840 pound of propane = 19168.4 KJ = 19168400 J

Temperature change of water 22⁰C to 46.8⁰C

T = 46.8 - 22 = 24.8⁰C

specific of water = 4.184 J/ g K

mass of water = ?

q = mass of water X specific heat of H₂O_(l) X T

19168400 = mass of water X 4.184 X24.8

mass of water = 19168400 / (4.184 X24.8)

mass of water = 184732 gm = 184.732 kg of water heated.

184.732 kg of water heated.from 22⁰C to 46.8⁰C

queen_honey_blossom answered 1 year ago

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