00:01
Hi, so to solve for this, we'll equate the amount of heat lost by copper.
00:09
I'm just going to write cu and then the amount of heat gained by water.
00:16
So this will be negative mass, specific heat, change in temperature for copper.
00:21
And then we have on the right side, positive mass, specific heat, change in temperature for h2o.
00:29
And our unknown here is the initial temperature of copper.
00:32
So let's plug in the information that we have.
00:36
Mass of copper is 80 grams, specific heat capacity of copper is 0 .20 joules per gram degrees celsius.
00:42
The change in temperature, the final temperature of the system is 26 .9 minus the unknown, the initial temperature of copper.
00:50
And then we have here mass of water, 100 grams, specific heat capacity of water is 4 .184 joules per gram degrees celsius.
00:59
It's not given here, but this could be easily looked at in standard reference materials.
01:04
The change in temperature for water is 26 .9 minus the initial temperature of water sample, 22 .3 degrees celsius.
01:11
Cancel some units, cancel grams, grams, degrees celsius.
01:18
And calculating this, we'll get negative 16 joules per degree celsius and then 26 .9 minus the initial temperature of copper, 100 multiplied by 4 .184 times 26 .9 minus 22 .3.
01:35
That will give us 1924 .64 joules.
01:40
And then let's isolate our unknown to be 1924 .64 joules divided by negative 16 joules per degree celsius...