Posted by **lyd** on Saturday, November 22, 2008 at 10:04pm.

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 -
**Jeremy**, Saturday, November 22, 2008 at 11:26pm
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.

## Answer This Question

## Related Questions

- physics - A 1.2 kg block of ice is initially at a temperature of -5°C. (a) If 5....
- physics - A 1.3- block of ice is initially at a temperature of -3.5. If 6.6×105 ...
- physics - A 40-g block of ice is cooled to −71°C and is then added to 590...
- physics - A 40-g block of ice is cooled to −71°C and is then added to 590 ...
- physics - A 5-kg block of ice is at a temperature of -27 °C. How much heat must ...
- Physics - A 10 kg block of ice has a temperature of -13°C. The pressure is one ...
- had - A 0.106 kg ice cube is taken out of a freezer with temperature -16°C. It ...
- physics - 0.1kg of ice initially at a temperature of -10 degrees Celsius is ...
- Physics - I was having trouble with two different physics problems when studying...
- Physics - I was having trouble with two different physics problems when studying...

More Related Questions