Measuring Heat Capacity: Solving for Heat Transfer in Thermal Contact

[jlmac2001]

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 celsius. After a minute of so, the temperature of the contents of the cup is 24 degrees celsius. 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?

 

[HallsofIvy]

(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?

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.

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

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).

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

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

(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?

It is the heat capacity you just calculated in (c), of course!

 

[PrudensOptimus]

c is specific heat, which is equal to ΔQ/mΔT.

C is heat capacity, is equal to ΔQ/ΔT.