Question

What is the total amount of heat required (in kJ) to convert 192 g of ice at -17 oC into steam at 122 oC?

Sp. Heat for ice = 2.09 J/goC

Sp. Heat for water = 4.18 J/goC

Sp. Heat for steam = 1.84 J/goC

ΔHfus = 6.01 kJ/mol

ΔHvap = 40.67 kJ/mol

Answer 1

Given that;

Amount of ice = 192 g

Initial temperature = -17 oC

Final temperature = 122 oC

Sp. Heat for ice = 2.09 J/goC

Sp. Heat for water = 4.18 J/goC

Sp. Heat for steam = 1.84 J/goC

ΔHfus = 6.01 kJ/mol

ΔHvap = 40.67 kJ/mol

Step I:

Freezing point of ice= 0.0oC

So temperature change

Heating the ice:
∆T = 0 - (-17) = 17 C

We know that q = = M* cp *dT

Here m = mass of ice

cp = specific heat

dT = temperature change

q = 192 g x 2.09 J/g*C x 17 = 6821.76 J

step II:

Melting the ice:

in this process; Temperature not required, since it's a phase change from solid to liquid at constant temperature.

q = 192g/18g/mole x 6.01 kJ/mo * 1000 J / 1 kJ = 64106.67 J

step III:

Heating the water:
∆T = 100 - 0 = 100 C

q = 192g x 4.18 J/g*C x 100 C = 80256 J

Step IV:

Vaporizing the water:
similar in this process; Temperature not required, since it's a phase change from liquid to gas at constant temperature.

q = 192 g/18g/mole x 40.67 kJ/mol * 1000 J / 1kJ = 433813.33 J

step V:

Heating the steam :
∆T = 122 - 100 = 22 C

q = 192 g x 2.09 J/g*C x 22 C = 8828.16 J

q(total) = 6821.76 + 64106.67 + 80256 + 433813.33 + 8828.16 = 593825.92J

593825.92 J * 1 .00 Kj /1000 J

= 593.83 KJ