Question

At 1 atm, how much energy is required to heat 91.0 g of H2O (s) at-16.0 degrees C to H2O (g) at 141 degrees celsius?

Answer 1

Temperature change of ice -16⁰C to 0⁰C

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

= 91 2.03 16

q₁ = 2955.68 J

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 91 gm of ice to water without change in temperature require heat 91 6.01/18.01528 = 30.358 KJ

q₂ = 30.358 KJ = 30358 J

Temperature change of water 0⁰C to 100⁰C

T = 100 - 0 = 100⁰C

specific of water = 4.184 J/ g K

mass of water = 91 gm

q₃ = mass of water   specific heat of H₂O_(l)  T

= 91 4.184 100

q₃ = 38074.4 J

Phase change from water to vapour

Heat of vaporization of water = 40.65 KJ/mol

That mean to convert 1 mole of water to 1 mole of vapour without change in temperature require heat 40.65 KJ/mol

1 mole of water = 18.01528 gm

to convert 18.01528 gm of water to vapour without change in temperature require heat 40.65 KJ then to convert 91 gm of water to vapour without change in temperature require heat 91 40.65/18.01528 = 205.334 KJ

q₄ = 205.334 KJ = 205334 J

Temperature change of vapour 100⁰C to 141⁰C

specific heat of H₂O_(g) = 1.996 J / g K

q₅ = mass of vapour specific heat of H₂O_(g)T

= 91 1.996 41

q₅ = 7447.076 J

total q = q₁ + q₂ + q₃ + q₄ + q₅ = 2955.68+30358+38074.4+205334+7447.076 = 284169.156 J

total energy require for heat 91 gm H₂O from -16⁰C to 141⁰C = 284169.156 J = 284.169156 KJ