The constants for H2O are shown here: Specific heat of ice: sice=2.09 J/(g??C) S

Question

The constants for H2O are shown here: Specific heat of ice: sice=2.09 J/(g??C) Specific heat of liquid water: swater=4.18 J/(g??C) Enthalpy of fusion (H2O(s)?H2O(l)): ?Hfus=334 J/g Enthalpy of vaporization (H2O(l)?H2O(g)): ?Hvap=2250 J/g Part A How much heat energy, in kilojoules, is required to convert 39.0g of ice at ?18.0 ?C to water at 25.0 ?C ? Express your answer to three significant figures and include the appropriate units. Part B How long would it take for 1.50 mol of water at 100.0 ?C to be converted completely into steam if heat were added at a constant rate of 21.0J/s ? Express your answer to three significant figures and include the appropriate units

Explanation / Answer

part A

Three stages here: 1. Ice from -18.0C to 0.0C, 2. Ice to water at 0.0C, and finally water from 0.0 C to 25.0 C

1. Q = mc(Tf - Ti) = 39(2.09)(0.0 -18.0) = 1467.18 J

2. Enthalpy of fusion (ice to water or vice versa) is 334 J/g so 39 g x 334 = 13026 J

3. Q = mc(Tf - Ti) = 39(4.18)(25 - 0) = 4075.5 J

Total heat energy in kJ = 1467.18+13026+4075.5 = 18568.68 J = 18.568 kJ

partB

The mass of the water m is 1.50 mol x 18g/mol=27g
you need to use the formula Q=m x Enthalpy of vaporization to calculate the energy needeed to break up the ties among atoms: Q1=27g x 2250 J/g

Now you can calculate the time t with this formula: t=Q1 / 21J/s=27 x 2250 / 21=2892.857 secs

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