From the following data, calculate the total heat needed to convert 0.219 mol ice at -6.50°C to liquid water at 7.26°C.
Melting point at 1 atm 0°C
csolid 2.09J/g°C
cliquid 4.21J/g°C
ΔHfus 6.02 kJ/mol
2007-01-20 16:21:58 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
The calculation is carried out in 3 steps. m= 0.219 moles
a. Heating the ice to 0 deg C
b. Converting ice to water at 0 deg C
c. Heating water to 7.26 deg C
The heating steps follow the formua
Q = m x c(phase) x (Tfinal-Torig)
The melting follows the formula
Q = m x (heat of fusion/mole)
Just plug in the numbers and do the 3 additions.
2007-01-20 17:15:06 · answer #1 · answered by cattbarf 7 · 0⤊ 0⤋
Get ready!! ive only just started my anthralpy change in chemistry so this might be incredibly wrong but either way it looks good hehe!!
mxc(L)x(t2-t1) This will give us the anthralpy change of the heating change when it is a liquid..
working out the mass! water is H2O there fore 2X1 + 16 equals 18 which is the molcular mass..
mass -> number of moles X molcular mass
2.09 X 18.00 equals 3.94
Keep to the same sig figs!
this gives us the mass then plug it in along with the relevant specific heat capacity and the tempo change then add the final two
3.92*2.09*6.50-> 53.55KjKmol-1
3.92*4.21*7.26-> 120.49KjKmol-1
Total Enthalpy change 174.04KjKmol-1
2007-01-27 02:51:26 · answer #2 · answered by Minx 2 · 0⤊ 0⤋
a million ) 10 g steam (200 stages C) ------------->steam (one hundred stages C) -------------->water at one hundred stages C ----------> water at 0 stages C ---------> ice at 0 stages C -------> ice at - 50 stages C. -10 x100 x2.03 - 10 x 40.6 x1000 /18 - one hundred x 10 x 4.18 - 10 x 6020 /18 - 10 x 50 x 2.06 = - 12839 j it somewhat is 12.839 Kjoules.the specific heat temperature of water is taken as 4.18 joulesper gram kelvin.
2016-12-16 09:35:41 · answer #3 · answered by ? 4 · 0⤊ 0⤋
CaatBarf is absolutely right!
2007-01-27 00:40:36 · answer #4 · answered by Anonymous · 0⤊ 0⤋