posted by Anonymous on .
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:
q= ( 2.1 J g-1 °C-1)*(800g)*(20°C)=...
Ice to water
q=(6.02 kJ / mol)*(800g)*(1 mol/18g)=...
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)=...
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
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.