You have:

-----------------

Q = Delta UHTI + Delta UMI + Delta UHTW + Delta UVAPW + Delta UHTST

Delta UHTI = ( mI ) ( CvI ) ( TMP - TI1 )

Delta UHTI = ( 3.20 kg )( 0.490 kcal/kg - C ) [ ( 0.0 C ) - ( -20 C ) = 31.36 kcal

Delta UMI = ( mI ) ( Lf ) = ( 3.2 kg ) ( 79.8 kcal/kg ) = 255.36 kcal

Delta UHTW = ( mW ) ( CvW ) ( TBPW - TFPW )

Delta UHTW = ( 3.20 kg ) ( 1.00 kcal/kg - C ) ( 100.0 C - 0.0C ) = 320 kcal

Delta UVAPW = ( mW ) ( LvW ) = ( 3.20 kg ) ( 541 kcal/kg ) = 1731.2 kcal

Delta UHTSTM = ( mSTM ) ( CvSTM ) ( TSTM2 - TBPW )

Delta UHTSTM = ( 3.20 kg ) ( 0.477 kcal/kg - C ) ( 125 C - 100 C ) = 38.16 kcal

Q = 31.36 kcal + 255.36 kcal + 320 kcal + 1731.2 kcal + 38.16 kcal

Q = 2376.08 kcal ~ 2376 kcal <------------------

Q = ( 2376 kcal ) ( 4186.7 J/kcal ) = 9.948 x 10^6 J <-----------

Breaking this down into stages:

First raise the temp from -20.0 to 0 takes 3.20*20*.490 = 31.4 kcal

Now melt the ice 3.20*79.8 = 255 kcal

Now raise the temp to 100 3.20*1.00*100 = 320 kcal

Now boil the water 3.20*541 = 1731 kcal

Finally raise the steam temp to 125 3.20*25*.477 = 38.2 kcal

So, the amount of heat is: 31.4 + 255 + 320 +1731 + 38.2 = 2376 kcal or 9.94e+6 Joules