Posted by **Anonymous** on Monday, November 15, 2010 at 7:40pm.

How much energy does it take to convert 0.800 kg ice at -20.°C to steam at 250.°C?

Specific heat capacities:

ice, 2.1 J g-1 °C-1

liquid, 4.2 J g-1 °C-1

steam, 2.0 J g-1 °C-1

ΔHvap = 40.7 kJ/mol, ΔHfus = 6.02 kJ/mol

This is what I tried to do:

Heating ice

q= ( 2.1 J g-1 °C-1)*(800g)*(20°C)=...

Ice to water

q=(6.02 kJ / mol)*(800g)*(1 mol/18g)=...

Heating water

q=(4.2 J g-1 °C-1)*(800g) * (20)

Is the change in temperature still 20? This is where I am confused...

Water to steam

q=(40.7 kJ/mol)*(800g)*(1 mol/ 18g)=...

Heating steam

q=(2.0 J g-1 °C-1)*(800g)*(change in temp=???? 250-(-20)???)

Please can you check if I am doing this right and I need help with the change in temperature

- Chemistry -
**DrBob222**, Monday, November 15, 2010 at 10:09pm
Portions may be right but most steps leave out something. Here is how you do it.

q1 = heat to move T of ice from -20 to zero C.

q1 = mass ice x specific heat ice x (Tfinal-Tinitial). Tfinal = 0 and Tinitial = -20; therefore, Tfinal-Tinitial = 0 -(-20) = +20.

q2 = heat to melt the ice at zero C to liquid water at zero C.

q2 = mass ice x heat fusion.

q3 = heat to move T from zero C to 100 C.

q3 = mass water x specific heat water x 100 (that is Tfinal-Tinitial) or (100-0 = 100).

q4 = heat to convert from liquid water at 100 C to steam at 100 C.

q4 = mass water x delta Hvap.

q5 = heat to move T of steam at 100 C to steam at 250 C.

q5 = mass x specific heat steam x (Tfinal-Tinitial).

Total energy required is

qtotal = q1 + q2 + q3 + q4 + q5.

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