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 (Tfinal-Tinitial). 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 remains--perhaps 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 (Tfinal-Tinitial) 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?

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