Heat capacity of a coffee cup calorimeter.

[dsterlingwells]

How do you solve for the heat capacity of a calorimeter?

[ATP]

How do you solve for the heat capacity of a calorimeter?

Post a question! ....

[ATP]

example question**

[dsterlingwells]

................Hot Water.........Cold Water
Volume........100mL ............100mL
Start Temp.....90 C................20 C
End Temp......54 C................54 C

What is the heat capacity of the students coffee cup?

[chad]

For heat capacity look at the unit...should be in J/degree C or cal/degree C (or temperature change in units of Kelvin...same diff). So the heat capacity of a calorimeter is usally calculated by C = q/(delta T).

This is a super tricky question fyi. If you calculate it out (q=mc(delta T)) using the specific heat of water to be 1 cal/(g(degree C))

then the hot water gave off 3600cal [q = (100)(1)(-36)]
whereas the cold water absorbed 3400cal [q = (100)(1)(34)].

This leaves 200cal unaccounted for; this must be what the coffee cup itself absorbed (assuming no heat lost to the surroundings). Assuming the coffee cup also started at 20 degrees C and ended up at 54 degrees C:

C = q/(delta T) = 200cal/(34 degrees C) ~ 5.9 cal/(degree C)

Hope this helps!

[dsterlingwells]

Yeah i couldn't find a simplified equation for the capacity of the calorimeter. The Kaplan book gives this long explanation but i couldn't understand it. The answer was in J/degree.. so it was 24.6..which goes with your answer.

thanks!