Why is specific heat capacity less?

[donkeycopter]

Hi,

I just did a specific heat capacity test using a calorimeter, and the specific heat capacity I found (using heat gained = heat lost) is less than the real value.

The problem is, I expected it to be more.

I mean heat will be lost through the calorimeter, meaning that a greater amount of heat energy will be required than ideally yes? So if the mass and the temperature are the same as in the ideal scenario in the equation:

Q = mc $$\Delta$$ TThen doesn't that mean heat energy required is directly proportional to c? So if heat energy required goes up (as heat is lost) doesn't that mean the specific heat will go up?I can't figure it out!

Thanks!

 

[Stickey]

How good is your calorimeter?

It should know, and compensate for, it's own heat loss

 

[donkeycopter]

Oh, it's an old manual calorimeter in which you just stick a thermometer in the whole in the lid.

 

[Stickey]

Then that's probably your reason! :D

A bit of investigation into the errors might explain it, and put the expected result within your error bars.

How far out were you? Did you repeat it and get the same thing?

 

[donkeycopter]

But that's not the point.

The textbooks says the specific heat capacity will be LESS, because heat is lost through the calorimeter, but I don't understand WHY.

 

[Curl]

You screwed up in measuring energy transfer IN, or the mass of the stuff you were testing.

Or your thermometer sucks.

No cause for alarm.

 

[nasu]

You calculate c as
c=Q/(m*delta T), right?
The problem is to find Q.
You assume that all the heat lost by the object you are measuring is absorbed by the water and calorimeter, Q = Qw+Qc
If some heat is lost to the environment too, then Q= Qw+Qc+Qe
If you have all these three terms and add them together, then plug into first formula, you'll get the right value for c. If you neglect Qe (you cannot measure it), then you'll use Q'=Qw+Qc which is less than Q and it will give a smaller value for c.