In order to determine the heat capacity of an unknown liquid...

In order to determine the heat capacity of an unknown liquid a student heats a 10.29-gram sample of cobalt to 92.21\deg C and places in into 38.40 grams of the unknown liquid that is at 17.58\deg C. The temperature of the system is 21.01\deg C once thermal equilibrium has been established. The heat capacity of cobalt is 0.421 J/g·\deg C. What is the heat capacity of the unknown liquid?

Transcript

00:01 Hi, so to solve for this, we will equate the amount of heat lost by the metal and the amount of heat gained by the water.

00:08 So this is heat lost by the metal since the cobalt sample is hotter compared to the water sample.

00:19 This will be negative mass specific heat change in temperature and this is for the metal.

00:26 And then we have here positive mass specific heat change in temperature for h2o.

00:32 Rearranging this to solve for the specific heat capacity of cobalt, we'll have positive mass specific heat change in temperature for h2o over negative mass and the change in temperature for cobalt.

00:53 Now let's plug in the information that we have.

00:57 Mass of the water sample is 50 grams.

01:01 Specific heat is 4 .18 joules per gram kelvin.

01:05 And let's convert the final temperature first.

01:09 It's 25 degrees celsius.

01:11 To convert this to kelvin, we have to add 2 sub -entity and this will give us 298.

01:17 Initial temperature for h2o sample is 20 plus 2 sub -entity.

01:23 This is 293.

01:26 And for the metal, initial temperature is 100.

01:31 Add 2 sub -entity.

01:32 This is the conversion factor.

01:33 So we have 3 sub -entity kelvin now.