Question

A.) How much heat energy, in kilojoules, is required to convert 60.0 g of ice at -18 degrees celcius to water at 25 degrees celcius?

B.) How long would it take for 1.50 mol of water at 100 degrees Celsius to be converted completely into steam if heat were added at a constant rate of 24.0 J/s?

Answer 1

A)

Temperature change of ice -18⁰C to 0⁰C

T = 18⁰C

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

= 60 X2.03 X 18

q₁ = 2192.4 J = 2.1924 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 60 gm of ice to water without change in temperature require heat 60 X6.01/18.01528 = 20.0163 KJ

q₂ = 20.0163 KJ

Temperature change of water 0⁰C to 25⁰C

T = 25⁰C

specific of water = 4.184 J/ g K

mass of water = 60 gm

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

= 60 X  4.184 X 25

q₃ = 6276 J = 6.276 KJ

total energy required = q₁ + q₂ + q₃

= 2.1924 + 20.0163 + 6.276 = 28.4847 KJ

heat energy required to convert 60.0 g of ice at -18 degrees celcius to water at 25 degrees celcius = 28.4847 KJ

B) first calculate energy required to convert 1.50 mole water at 100⁰C to steam

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

then to convert 1.5 mole of water required heat = 1.5 X 40.79 = 61.185 KJ = 61185 J

Heat added at constant rate 24.0 j/s

then to add 61185 J heat required time = 61185 / 24 = 2549 second

it take for 1.50 mol of water at 100 degrees Celsius to be converted completely into steam if heat were added at a constant rate of 24.0 J/s = 2549 second = 42.48 minute