00:01
Hi, so to solve for this, we will equate the amount of heat lost by copper and then the amount of heat gained by water.
00:08
So heat lost by copper since copper is compared to, is hotter compared to the water sample.
00:15
So the heat gained by h2o.
00:20
On the left side of the equation, we'll have negative since this is heat lost, mass is specific heat and change in temperature.
00:26
This is for copper and then we'll have positive mass specific heat change in temperature of water on the right side of the equation.
00:35
Our unknown here is the mass of water in the container.
00:39
So that means let's rearrange this equation.
00:42
Mass of water is therefore equivalent to negative mass specific heat change in temperature.
00:49
This is for copper divided by the specific heat capacity of water and then the change in temperature.
01:01
Now let's plug in the information that we have.
01:06
Mass of our sample copper is 65 grams.
01:10
Specific heat of copper is given here 0 .20 joules per gram degree celsius.
01:17
The change in temperature, final temperature of the system is 24 .9 minus initial temperature of copper 97 .6 degrees celsius...