Part 1: You heat 64.2 grams of water from -24.4 oC to 138.2 oC and obtain the he

Question

Part 1:

You heat 64.2 grams of water from -24.4 oC to 138.2 oC and obtain the heating curve given in the introduction. What phase(s) is/are present in the line between B and C?

Notes:

Water's Boiling Point: 100 oC

Water's Melting Point 0 oC

Cice =2.06 J/(g*oC)

Cwater = 4.184 J/(g*oC)

Cwater vapor = 2.02 J/(g*oC)

Hvap = 40.65 kJ/mol

Hfus = 6.02 kJ/mol

You will use the following heating curve to answer this question:

Part 3:

a) How many kJ of energy are necessary to raise the ice to the melting point (but not to melt the ice)?

b) After you have heated the ice to the melting point, how many additional kJ of energy are necessary to melt the ice?

c) After the ice is melted, how many additional kJ of energy are needed to raise the water to the boiling temperature?

d) After the water is raised to the boiling temperature, how many additional kJ of energy are needed to boil the water?

e) After the water is boiled, how many additional kJ of energy are needed heat the water vapor to the end temperature?

f) How many TOTAL kJ of energy were used in the process described in part 1?

Explanation / Answer

A to B is heating of ice phase.

B to C is melting of ice.

C to D is heating of water.

D to E is vaporisation of water to steam.

E to F is heating of steam.

so, from B to C, both solid (ice) and liquid(water) are coexisting.

a) We have 64.2 gram of ice at -24.4° C

So energy needed to raise it to melting point i.e. 0°C=> Cice * T * m

=> 64.2 g* 2.06 J/°C g * (24.4-0)°C

=> 3226.95 J => 3.23 kJ

b) Now, we have ice at 0° C.

ice Molecular weight=> 18g/mol

so, 64.2 g=> 64.2 g/ 18g/mol=> 3.57 moles of ice

to melt it, energy required=> Hfus* moles.of ice

=> 6.02 kJ/mol* 3.57 moles

=> 21.49 kJ required for melting the ice

c) A Now we have 64.2 grams of water at 0°C.

boiling point is 100°C.

so, heat required=> m* Cwater* T

=> 64.2 g* 4.184 J/g°C * (100-0)°C

=> 26861 J=> 26.86 kJ

d) Now similar to melting,

energy required for boiling => moles of water* Hvap=> 3.57 moles* 40.65 kJ/mol=> 145.12 kJ

e) After water is boiled, energy needed to heat the steam to 138.2 °C

=> m* cwater vapor * T

=> 64.2 g * 2.02 J/g°C * (138.2-100)

=> 4953 J => 4.95 kJ

f) total energy used=> 21.49 kj+ 3.23 kJ + 26.86 kJ + 145.12 kJ + 4.95 kj=> 201.65 kJ

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