Part A Calculate the enthalpy change, H , for the process in which 24.5 g of wat

Question

Part A

Calculate the enthalpy change, H, for the process in which 24.5 g of water is converted from liquid at 5.0 Cto vapor at 25.0 C .

For water, Hvap = 44.0 kJ/mol at 25.0 C and Cs = 4.18  J/(gC) for H2O(l).

Express your answer to three significant figures and include the appropriate units.

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Part B

How many grams of ice at -9.0 C can be completely converted to liquid at 26.4 C if the available heat for this process is 4.17×103 kJ ?

For ice, use a specific heat of 2.01 J/(gC) and Hfus=6.01kJ/mol.

Express your answer to three significant figures and include the appropriate units.

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Explanation / Answer

Q1) The process is

H2O(l) -------------------------> H2O(l) --------------------> H2O(g)

5C q1 25C q2 25C

Total heat absorbed = q1 + q2

q1 = mass x specific heat x delta T

= 24.5g x 4.18J/g.C x 20C =2048.2 J

q2 = moles x delta Hvap

= 24.5 x 44.0x1000J /18

= 59888.88 J

thus total heat = 2048.2 +59888.88 J

=61937.08 J

=61.937 kJ

Q2)

the process is

H2O(s) ------------> H2O(s) -------------> H2O(l) -------------> H2O(l)

-9C q1 0C q2 0C q3 26.4C

Total heat = q1 +q2 +q3

= 4.17x103 kJ = 4170 J

Let mass of water = m g

q1= mass x specific heat of ice x delta T

= m gx 2.01 J /g.C x 9 C

= 18.09 mJ

q2 = mass x delta H fusion

= m g x 6.01 x1000 /18 J/g

= 333.88 m J

q3 = mass x specific heat of water x delta T

= m g x 4.18J/g.C x 26.4 C

= 110.352 m J

Thus total heat = 4.17 x103 kJ = 4170kJ

= 4170 x1000 J

=18.09 mJ + 333.88 m J +110.352 m J

=462.32 mJ

equating

4170 x1000 J = 462.32 xm J

Thus m = 9019.72 g

Mass of ice = 9019.72 g

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