Posted by lyd on Saturday, November 22, 2008 at 10:06pm.
A 1.2 kg block of ice is initially at a temperature of 5°C.
(a) If 5.8 multiplied by 105 J of heat are added to the ice, what is the final temperature of the system?
°C
(b) Suppose the amount of heat added to the ice block is increased by a factor of 3.0. By what factor must the mass of the ice be increased if the system is to have the same final temperature?
multiplied by

physics  DrBob222, Saturday, November 22, 2008 at 10:15pm
You must do the (a) part in stages.
q1 = heat required to move the ice from a T of 5 to zero.
mass x specific heat ice x (TfinalTinitial). You have mass, look up specific heat ice and delta T = 5.
q2 = heat to melt the ice.
q2 = mass x heat fusion.
You have mass. Look up heat of fusion.
Now q1 + q2 have been used. Subtract from the 5.8E5 J to determine the amount of heat remaining (if any remainsperhaps not all of the ice will melt) to raise the T of water from zero to unknown T.
q = massH2O x specific heat H2O x (TfinalTinitial) and solve for Tfinal. Tinitial is zero. Post your work if you need further assistance. 
physics  lyd, Saturday, November 22, 2008 at 10:38pm
Q1= 1.2 x 2.05 x 5 = 12.3
q2= 1.2 x 333 = 399.6
q1 + q2 = 411.9
580000  411.9 = 579588.1
what did i do wrong?