Which takes more energy to heat up: air or water? How about water versus metal or water versus another liquid like soda?

These questions and many others are related to a property of matter called specific heat. Specific heat is the amount of heat per unit of mass needed to raise a substance's temperature by one degree Celsius.

So it takes more energy to heat up water than air because water and air have different specific heats.

#### TL;DR (Too Long; Didn't Read)

Use the formula:

**Q = mcΔT**, also written **Q = mc(T - t _{0})**

to find the initial temperature (t_{0}) in a specific heat problem.

In fact, water has one of the highest specific heats of any "common" substance: It's 4.186 joule/gram °C. That's why water is so useful in moderating the temperature of machinery, human bodies and even the planet.

## Equation for Specific Heat

You can use the property of specific heat to find a substance's initial temperature. The equation for specific heat is usually written:

**Q = mcΔT**

where Q is the amount of heat energy added, m is the substance's mass, c is specific heat, a constant, and ΔT means "change in temperature."

Make sure your units of measurement match the units used in the specific heat constant! For example, sometimes the specific heat may use Celsius. Other times, you'll get the SI unit for temperature, which is Kelvin. In these cases, the units for specific heat will either be Joules/gram °C or else Joules/gram K. The same could happen with grams versus kilograms for the mass, or Joules to Bmu for energy. Be sure to check the units and make any conversions needed before you get started.

## Using Specific Heat to Find Initial Temperature

ΔT can also be written (T - t_{0}), or a substance's new temperature minus its initial temperature. So another way to write the equation for specific heat is:

**Q = mc(T - t _{0})**

So this rewritten form of the equation makes it simple to find initial temperature. You can plug in all the other values that you're given, then solve for t_{0}.

For example: Say you add 75.0 Joules of energy to 2.0 grams of water, raising its temperature to 87 °C. Water's specific heat is 4.184 ^{} Joules/gram °C. What was the initial temperature of the water?

Plug the given values into your equation:

75.o J = 2.0 g x (4.184 J/g°C) x (87 °C - t_{0}).

Simplify:

75.o J = 8.368 J/°C x (87 °C - t_{0}).

8.96 °C = (87 °C - t_{0})

78°C = t_{0}.

## Specific Heat and Phase Changes

There's one important exception to keep in mind. The specific heat equation doesn't work during a phase change, for example, from a liquid to a gas or a solid to a liquid. That's because all the extra energy that's being pumped in is being used for the phase change, not for increasing the temperature. So the temperature stays flat during that period, throwing off the relationship between energy, temperature and specific heat in that situation.

References

About the Author

Elise Hansen is a journalist and writer with a special interest in math and science.