The heat of fusion of ice is 333 J/g (meaning 333 J is absorbed when 1 gram of i

Question

The heat of fusion of ice is 333 J/g (meaning 333 J is absorbed when 1 gram of ice melts). The heat of vaporization of liquid water at 100 degree C is 2260 J/g. a. Calculate the molar enthalpy change in enthalpy, Delta H, for these two processes. H_2 O_(s) rightarrow H_2 O_(i) Delta H = ? H_2 O_(i) rightarrow H_2 O_(g) Delta H = ? b. Determine the number of grams of a 10.0 g sample of ice at an initial temperature of -5.0 degree C that are converted to liquid water following the addition of 0.800 kJ of heat.

Explanation / Answer

a.) Did you notice that the heats of fusion and vaporization were given in joules and not kilojoules? Using the periodic table, we know that 1 mole of water (H2O) is 18.02 g. Therefore:

fusion H = 18.02 g x 333 J / 1 g
fusion H = 6.00 x 103 J
fusion H = 6.00 kJ

vaporization H = 18.02 g x 2257 J / 1 g
vaporization H = 4.07 x 104 J
vaporization H = 40.7 kJ

So, the completed thermochemical reactions are:

H2O(s) H2O(l); H = +6.00 kJ
H2O(l) H2O(g); H = +40.7 kJ



b.) Now we know that:

1 mol H2O(s) = 18.02 g H2O(s) ~ 6.00 kJ

Using this conversion factor:

0.800 kJ x 10 g ice / 6.00 kJ = 1.33 g ice melted

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