PLEASE ANSWER THIS QUESTION AND TeLL ME HOW YOU GOT YOUR ANSWER?

Question

what is the final temperature when a 3kg gold bar at 99 degrees celsius is dropped into .22kg of water at 25 degrees celsius?

Answer 1

You'll have to find or look up the specific heat of gold. Once you have that, you can use the following equation:

q(water) = -q(gold)
220(4.184)(T2-25) = -3000 c (T2-99) Where c = the specific heat of gold.

You'll just have to solve for T2, and you've got it.

Answer 2

ASSUMING the gold bar and water are in a PERFECTLY insulated container, and there is absolutely NO heat loss, the easiest way is to set up a formula thusly;

3.0 kg @ 99 degrees provides 3 x 99 = 297 "kg-degree"
0.22kg @ 25 degrees provides 0.22 x 25 = 5.5 "kg-degree"

Then divide total "kg-degree" of 302.5 by total weight of 3.22kg to get 93.944 degrees as final temperature

Unfortunately, this doesn't work in real life because no container is perfectly insulated - however, hopefully, you get the idea?