# Simple question about specific heat

1. Aug 26, 2006

### Signifier

Say I have a metal and I heat it up to some temperature T. Then I dunk it in isolated water of temperature t, water which is fairly cool (t << T). Eventually, the metal and the water are at some final temperature Tf. The specific heat of the substance is calculated by:

c = (Mass of water)(Specific heat of water)(Change in water temp) / (Mass of metal)(Change in metal temp)

I perform the experiment a bunch of times and consistently get about the same specific heat c. Now, what would happen if halfway through the cooling process (where the water and the metal are becoming the same temperature), I poured some boiling water into the container holding the water and the metal? What would happen to the calculated specific heat? Would it drop or raise? Or would it stay the same? And why?

Thanks!

2. Aug 26, 2006

### Farsight

Signifier: this question is intended to make you get the hang of "specific heat" as distinct from heat or temperature. I'd like to tell you an answer, but I think it's best if you look it up on google and work it out for yourself.

3. Aug 26, 2006

### Signifier

Zing... if only this were a homework question (I posted it here because I didn't want to bother the more advanced physics boards). Nevertheless, my first thought is... c will go down. However, to get this result I assumed that the final temperature that the water and the metal reach (IE, when the temperature of the two "stops changing" essentially) is a constant that does not vary as the initial temperature of the water varies. Assuming this and using conservation of energy, measured c would decrease as the initial temperature of the water increases and everything else is held constant.

Still not certain here... any tips would be greatly appreciated.

4. Aug 27, 2006

### Andrew Mason

The specific heat is a property of the metal. It has some temperature dependence but I don't think it is very much in this temperature range. In making your measurement, you would have to take into account the change in temperature of all the water and the metal.

AM

5. Aug 27, 2006

### Farsight

The calculated specific heat stays the same. That's assuming the setup is jacketed so it doesn't lose dramatic amounts of heat to the outside world. Suppose you started with a litre of icy water at 0oC, and it was on the way to 50oC. Then you add another litre of water at 100oC. It's like starting with 2 litres of water at an intermediate temperature and somewhat cooler metal.

6. Aug 28, 2006

### loom91

There's an wonderful formula, known as the fundamental principle of calorimetry, which can be used to do virtually all sums in calorimetry. It says that if there be a set of substances (in any physical state) $J_1, J_2, ..., J_n$ with masses $m_1, m_2, ..., m_n$ and specific heats $s_1, s_2, ..., s_n$, and all these are brought into complete thermal contact with initial temparatures $$t_1, t_2, ..., t_n$$, then the resulting common temparature in thermal equilibrium is given by

$$t_c = \frac {\sum_i m_is_it_i}{\sum_i m_is_i}$$

In other words, it is the mean of the initial temparatures weighted by their respective heat capacities. Plug in all values into this wonderful little equation and all results will come popping out. Never try to do physics qualitatively.

Last edited: Aug 28, 2006