Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Measuring Heat capacity

  1. Jan 26, 2004 #1
    Here's the problem:

    To measure the heat capacity of an object, all you usually have to do is put it in thermal contact wit another object whose heat capacity you know. As an example, suppose that a chuck of metais immersed in boiling water (100 degrees), then is quickly transferred into a Styrofoam cup containing 250 g of water at 20 degrees Celcius. After a minute of so, the temperature of the contents of the cup is 24 degrees celcius. Assume during this time n significant energy is transferred between the contents of the cup and the surroundings. The heat capacity of the cup itself is negligible.

    (a) How much heat gained by the water?

    For this one, would I use C = Q/delta T and solve for Q? If so, what is C?

    (b) How much heat is lost by the metal?

    Not sure about this one? Don't know how to start this one.

    (c)What is the heat capacity of this chunk of metal?

    Would I use the same eqn. that I used in part a?

    (d)If the mass of the chunk of metal is 100g, what is its specific heat capacity?

    I think I would use c=C/m. But what is C?
     
  2. jcsd
  3. Jan 27, 2004 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, assuming that "delta T" is "change in T" (I had first thought of T as time!) delta Q= C* delta T (Notice that I am saying "delta Q" rather than Q). Notice that the problem said "put it in thermal contact with another object whose heat capacity you know". What is the heat capacity of water? The water has increased temperature from 20 to 24 degrees: it's heat content has increased by 4 times the heat capacity of water.

    Where did any heat lost by the metal go? You are told to " Assume during this time no significant energy is transferred between the contents of the cup and the surroundings. The heat capacity of the cup itself is negligible." Any heat lost by the metal is gained by the water. This answer should be exactly the same as the answer to (a).

    Yes. The metal has decreased tempeature from 100 degrees to 24 degrees so delta T is 76. Now that you know delta Q (from (a) or (b)) and delta T, you can solve for C.

    It is the heat capacity you just calculated in (c), of course!
     
  4. Jan 27, 2004 #3
    c is specific heat, which is equal to ΔQ/mΔT.

    C is heat capacity, is equal to ΔQ/ΔT.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Measuring Heat capacity
  1. Heat Capacity (Replies: 3)

  2. Specific heat capacity (Replies: 4)

Loading...