Question

calculate the minimum number of grams of propane, C3H8(g), that must be combusted to provide the energy necessary to convert 3.40 kg of ice at -16.0 ∘C to liquid water at 79.0 ∘C.

Answer 1

First calculate energy necessary to convert 3.40 gm of ice at -16⁰C to liquied water at 79.0⁰C

Temperature change of ice -16⁰C to 0⁰C

T = 16⁰C

mass of ice = 3.40 kg = 3400 gm

specific heat of H₂O_(s) = 2.03 J /gm⁰C

q₁ = mass of ice X   specific heat of H₂O_(s) X T

= 3400 X2.03 X 16

q₁ = 110432 J = 110.432 KJ

Phase change from ice to water

Heat of fusion of water = 6.01 KJ/mol

That mean to convert 1 mole of ice to water without change in temperature require heat 6.01 KJ/mol

1 mole of water = 18.01528 gm

To convert 18.01528 gm of ice to water without change in temperature require heat 6.01 KJ then to convert 3400 gm of ice to water without change in temperature require heat 3400 X6.01/18.01528 = 1134.259 KJ

q₂ = 1134.259 KJ

Temperature change of water 0⁰C to 79⁰C

T = 79⁰C

specific of water = 4.184 J/ g ⁰C

mass of water = 3400 gm

q₃ = mass of water Xspecific heat of H₂O_(l) X T

= 3400 X 4.184 X 79

q₃ = 1123822 J = 1123.822 KJ

q energy requied = q₁ + q₂+ q₃ = 110.432 KJ + 1134.259 KJ + 1123.822 KJ = 2368.5 KJ

heat energy necessary to convert 3.40 gm of ice at -16⁰C to liquied water at 79.0⁰C = 2368.5 KJ

heat of combution of propane = 2220 KJ/mole

1 mole propene produce 2220 KJ energy then to produce 2368.5 KJ energy requiered propane = 2368.5 /2220 =

1.067 mole of propane

molar mass of propane = 44.1 gm/mol

then 1.067 mole propane = 1.067 X 44.1 = 47.05 gm

47.05 of propane must be combusted