Question

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

Answer 1

heat = mCpTsub(f-i)

609 J = (1.2kg)(4184 J/kgC)(Tf - Ti)

[609 J/ [(1.2kg)(4184 J/kgC)]] -5C = Tf

Tf = -4.87... degrees C. for part A

For Part B:

Use the same equation, except multiply heat by 3. Solve for the new mass, then divide that mass by the old mass (1.2kg) to get the factor. It's probably 3, but I don't feel like checking! Good luck.