Question

Some physical properties of water are shown below:
melting point 0.0°C
boiling point 100.0°C
specific heat solid 2.05 J/g·°C
specfic heat liquid 4.184 J/g·°C
specific heat gas 2.02 J/g·°C
ΔH° fusion 6.02 kJ/mol
ΔH° vaporization 40.7 kJ/mol


80.0 kJ of heat are added to 20.5 g of ice at -24.3°C.

(a) What will be the final temperature of the system (in °C)?
(b) What mass will be in each phase after the heating?
i.solid - 0
ii. liquid - 0
iii. gas -


I need help with a and with b.iii

My Work:
q = m * c * (change in temperature)
q = 20.5 * 2.05 * (0+24.3) = 1021.2 J
q = 20.5 * 4.184 * (100-0) = 8577.2 J

80,000 J - (1021.2+8577.2) = 70401.6 J left for the final phase

How do I find both the final mass and final temperature if they are both variables?

Answer 1

q = m * c * (change in temperature)
q = 20.5 * 2.05 * (0+24.3) = 1021.2 J <b> right and I would make that 1.02 kJ.

Then you need to melt the ice.
q = (20.5/18)*6.02 = ?</b>

q = 20.5 * 4.184 * (100-0) = 8577.2 J <b> right and I would make that 8.68 kJ.

Then you need to vaporize the water.
(20.5/18) x 40.7 = ? kJ.

80,000 J - (1021.2+8577.2) = 70401.6 J left for the final phase <b>Not right since I've added some steps you left out.

80 - each of the above = ?kJ left and I would change that back to J.

Then q(in J) = 20.5 x specific heat steam x (Tfinal - Tinitial).
Tf is the unknown. Ti is 100 C</b></b>