Question

How do you calculate specific heat?

Answer 1

In what context? From experimental data? That's a question of data analysis, and basically you multiply and divide and add and subtract measurements, depending on how you measure it.

From first principles? That's a challenge in statistical mechanics, and you begin with a microscopic model of the motion of atoms and molecules, and then dive into either hairy math or massive numerical calculation.

Answer 2

Here's my problem:
A 20.94g sample of an unknown metal is heated to 99.4 degrees C in a hot water bath until thermal equilibrium is reached. the metal is quickly transfered to 100ml of water at 22.0 degrees C contained styrofoam cup. the thermal equilibrium temperature of the metal plus water mixture is 24.6 degrees C. what is the specific heat of the metal?"

Answer 3

Basically, I can't figure out how to set up the equation. I can do the math, I just don't know what to calculate, and google searches prove fruitless.

Answer 4

The heat removed from the hot metal by the cold bath would be equal to its specific heat capacity multiplied by its mass and the change in temperature, yes? You're not unclear on what heat capacity is in the first place?

So write an equation for that statement, putting in values you know (e.g. the change in temperature and mass) and assigning variables to things you don't know, like the heat capacity and heat loss. By my reckoning you've now got one equation with two unknowns, which is not by itself soluble. You need at least one more equation.

So let's look at the bath. Here, you also know the heat gained by the bath equals its heat capacity multiplied by its mass and the change in temperature. You also know, courtesy of the Law of Conservation of Energy, that the heat gained by the bath must equal the heat lost by the hot metal. So write another equation for the heat gained by the bath, again putting in the values you know, and assigning variables to things you don't know. (You are probably expected to know, or look up, the specific heat capacity of water.)

It seems to me that you will then have two equations, and still only two unknowns, the heat transferred and the heat capacity of the metal. But you can solve two equatios in two unknowns. So now it's just algebra.

Answer 5

As a general rule, I would say the most common "blindness" of students solving complex problems is to not realize that you may be expected to begin with more than one equation -- that you will not always be able to leap in one act of inspiration to a single equation with the desired quantity on the left hand side and nothing but numbers on the right.

So a more fruitful approach, I think, to word problems, is to relax and not worry about having one single equation at first. Just start writing down all the facts and relationships you realize are true about the situation, and the data you're given. Write down *lots* of equations, and define variables all over the place freely. Then see if you have the same number of unknowns as you do equations. If so -- you're home free! If not...then you need to start looking harder, trying to find new equations. Sometimes these are significant -- applications of some physical law equation or other -- and sometimes they're very simple, like realizing two variables you'd assigned are in fact equal to each other, or one is the negative of the other, or twice the other, et cetera.