We've discussed the concepts of heat, temperature, and specific heat. Now, it's time to put them all together mathematically. This is the heat transfer equation:

**q = CmΔT**

Where:

**q** is the heat in Joules**C** is the specific heat in Joules per gram per degree Kelvin**m** is the mass of the object in grams**ΔT** is the change of temperature in degrees Kelvin.

### Sample Problem

How much heat is required to raise the temperature of a kilogram of water by ten degrees Kelvin?

We take the above equation, plug 1,000 grams in for m and 10 Kelvin in for ΔT. But what do we plug in for C? We have to find the specific heat of water.

Find a table of specific heat capacities online, and you'll see that water has a value of 4.18 Joules/gK. Plug that value in, solve the equation, and you get:

q = 41,800 Joules.

### Sample Problem Two

Now, consider a different problem:

How much will the temperature of a kilogram of silver increase if you add 41,800 Joules (the same amount of heat and mass as in the problem above)?

Let's first rearrange our equation to solve for ΔT. We get:

**ΔT = q/(Cm)**

Then, we look up the specific heat of silver (it's .24) and solve for ΔT. We get:

ΔT = 174 degrees Kelvin

Yikes, that's hot! It's much easier to raise the temperature of silver than water.

**Water has an exceptionally high specific heat capacity. That's why it does such a good job of cooling you down.**

If you tried to use a piece of cold silver to cool you down, it would quickly reach your skin temperature and stop transferring heat.

But water will absorb a lot more heat before it reaches your skin temperature (if it ever does).