Question

Calculate the amount of energy (in kJ) necessary to convert 317 g of liquid water from 0°C to water vapor at 187°C. The molar heat of vaporization (Hvap) of water is 40.79 kJ/mol. The specific heat for water is 4.184 J/g · °C, and for steam is 1.99 J/g · °C. (Assume that the specific heat values do not change over the range of temperatures in the problem.)

Answer 1

Temperature change of water 0⁰C to 100⁰C

T = 100⁰C

specific of water = 4.184 J/ g ⁰C

mass of water = 317 gm

q₁ = mass of water X  specific heat of H₂O_(l) X T

= 317 X 4.184 X 100

q₁ = 132632.8 J = 132.6 KJ

Phase change from water to vapour

Heat of vaporization of water = 40.79 KJ/mol

That mean to convert 1 mole of water to 1 mole of vapour without change in temperature require heat 40.79 KJ/mol

1 mole of water = 18.01528 gm

to convert 18.01528 gm of water to vapour without change in temperature require heat 40.79 KJ then to convert 317 gm of water to vapour without change in temperature require heat 317 X 40.79/18.01528 = 717.75 KJ

q₂ = 717.75 KJ   

Temperature change of vapour 100⁰C to 187⁰C

q₃ = mass of vapour Xspecific heat of H₂O_(g) X T

= 317 X 1.99 X 87

q₃ = 54882.21 J = 54.88KJ

total heat required to convert 317 gm ice fron 0⁰C to 187⁰C = q₁ + q₂ + q₃

= 132.6 + 717.75 + 54.88 = 905.23 KJ

total heat required to convert 317 gm ice fron 0⁰C to 187⁰C = 905.23 KJ