How To Calculate Specific Heat

What is specific heat?

The specific heat (also called specific heat capacity) is an important property of materials because it tells us how much energy will be needed to heat or cool a certain substance. With that information we can know long the cooling or heating process will take, and how much it will cost.

Metals for example have a low specific heat, which means only a small amount of heat is needed to raise their temperature. Iron has a lower specific heat of than water. This is why it takes much more energy to heat up water than metals. This can be easily seen when we boil water. After turning the stove on, it can be seen that the pot itself gets hot very quickly, much faster than the water in it.

This article will provide the specific heat equation, the steps to calculate it and a few examples on how it can be applied.

Specific Heat Formula

Specific heat definition: In thermodynamics, the specific heat c_p of a substance is the heat capacity of a sample of the substance divided by the mass of the sample. In other words, it is the amount of heat energy that must be added to 1kg of the substance in order to cause an increase of 1 ºC.

Here c_p is the specific heat capacity, Q is the heat energy, m is the mass and ΔT is the change in temperature.

How To Calculate Specific Heat

Step 1: Convert the heat energy to units of Jules [J].

Step 2: Convert the mass to units of kilograms [kg].

Step 3: Calculate the temperature difference by subtracting the largest temperature minus the lowest, because the result has to be a positive number.

Step 4: Calculate c_p with the formula above.

Examples

Example 1

Calculate the specific heat of water if we need 4200 J to change the temperature of 1kg by 1 ºC.

In this example we have:

Q = 4200 J

m = 1kg

∆T = 1 ºC

Applying the formula:

Interestingly enough, we get the same number (4200) as the heat energy Q that we are given. But that shouldn’t be a surprise, since the definition of the specific heat capacity is the amount of heat energy Q that is needed to change the temperature of 1kg by 1 degree, and those are exactly the values that we are given.

Example 2

Now we will calculate the specific heat of water again, but with other values for Q, m and ∆T. We should still get the same c_p, since the specific heat capacity is a property of the substance itself, and does not depend on how much of it we have or how much we heat it up.

Calculate the specific heat of water if we need 2.52 kJ to change the temperature of 300 g by 2 ºC.

In this example we have:

Q = 2.52 kJ = 2520 J

m = 300 g = 0.3 kg

∆T = 2 ºC

Applying the formula:

As we can see, we get the same answer as in Example 1, since the specific heat capacity of a substance (in this case water) does not change.

Example 3

Calculate the specific heat of an oil if we need 5040 J to change the temperature of 60 g by 40 K.

In this example we have:

Q = 5040 J

m = 60 g = 0.06 kg

∆T = 40 K = 40 ºC (This is because the change in temperature is the same for ºC and K. It is true that a certain temperature T, like 25 K, is not the same as 25 ºC. But the change of temperature is the same).

Applying he formula:

Example 4

You are also likely to encounter exercises where the heat capacity is given, and you have to calculate the heat energy Q.

How much heat is necessary to heat up 200 g of water from 25 ºC to 29 ºC? The specific heat capacity of water is 4200 J/(kg ºC)

In this example we have:

c_p = 4200 J/(kg ºC)

m = 200 g = 0.2 kg

∆T = 29 – 25 = 4 ºC (Remember that you always need to get a positive answer, so subtract the larger value minus the smaller one).

Because now we want to calculate the heat, we rearrange the formula as such:

How to find the specific heat capacity in a table

Very often the value for the heat capacity is not given, but you have to look it up on a table, like the one on the left. Here it is very important to beware of the units.

As you can see on the table, the middle column uses the units of kilogram (kg) and the last column the units of gram (g). Both values are expressing the same specific heat capacity, but with different units. In order to solve exercises correctly, you have to take care to choose the right units. So far, we have solved examples where the specific heat capacity has the units J/(kg ºC). But it could very well be that you are only given the units of J/(g ºC) . In that case there are two options:

Option 1: You convert the units of the specific heat capacity from J/(g ºC)  to  J/(kg ºC)  so that we can solve the exercises in the same way we did before.

Option 2: You can keep the units of J/(g ºC) but then you have to convert the mass to g. That makes sense, since the units of all the terms should be the same. Therefore, if we see grams in the denominator, we should use grams as the mass unit.

Example 5

How much heat is necessary to heat up 0.05 kg of Cobalt from 30 ºC to 40 ºC? The specific heat of Cobalt is 0.41868 J/(g ºC)

Option 1: We convert the units of c_p to J/(kg ºC)

Notice that we multiplied by 1000 because we want the units g to cancel out, and in the end, what remains are the units kg.

m = 0.05 kg

∆T = 40 – 30 = 10 ºC

Applying the formula:

Option 2: We keep the units of the specific heat but convert the mass to grams

c_p = 0.41868 J/(g ºC)

m = 0.05 kg = 50 g (We convert the mass to grams, since we also have grams in the specific heat)

∆T = 40 – 30 = 10 ºC

Applying the formula:

Notice that we get the same result, which makes sense: It doesn’t matter what units we use for the mass, since in the end they cancel out with the units of the specific heat. One just has to make sure that the units being used are consistent. Here is a table with a large list of the specific heat of various substances.

Things to keep in mind

1. Make sure the change in temperature ∆T is always positive. It does not matter if the temperature is in Kelvin or Celsius.

2. Don’t forget to convert all the values to the right units.

3. If you are given the specific heat in units of grams like c_p = 0.41868 J/(g ºC), you can either convert it in units of kilograms, or not change the units but instead convert them to g. What matters in the end is that you use the same units.

4. It could also be possible to be given a c_p with kJ instead of J as units. In that case you have to convert the units to J. Or you can keep the units in kJ, but then you answer will be in kJ.

5. You might also encounter the concept of heat capacity. This is different from specific heat capacity, but they are related. The word specific tells us that it’s divided by the mass. So, if you want to calculate the heat capacity of a substance, you simply multiply the specific heat capacity times the mass of the substance. The heat capacity is written with a capital letter as such: C_p

6. The heat capacity of a substance is constant, which means it does not depend on how much mass we have, how much heat we apply, or what the temperature difference is (in reality it does depend on the temperature, but we normally make this simplification. Here you can see an explanation).

Conclusion

The specific heat is a really important property of materials. Substances that have a small specific heat can be heated up quickly. They also experience a large raise in temperature with only a small amount of heat. This makes it useful for cooking instruments, like cooking pans, because the metal gets hot very quickly. Thermometers are also made from materials with a low specific heat, so that they heat up quickly even when a small amount of heat energy is supplied.

On the other hand, substances with high specific heat are used for materials that we don’t want to get hot too fast. Some examples are kettle handlers or insulators. The fact that water has a high specific heat means that it is able to retain much more heat without the temperature raising too much, which enables oceans to absorb energy and helps to stabilize Earth’s habitable conditions. This also enables warm-blooded animals, like us humans, to be able to maintain our body temperatures, since we are 60% water.