Specific Heat Capacity

Have you ever used an automatic dishwasher? When a dishwasher door is opened a few minutes after the end of the washing cycle, you will find the ceramics and the heavy metal items will be completely dry. However, anything made of plastic will still be wet. This happens because plastic has a relatively low specific heat capacity, which means that it does not retain as much heat as the other material items and hence is not able to evaporate off the water droplets as quickly. In this article, we will learn all about specific heat capacity and investigate this property in different materials!

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Jetzt kostenlos anmeldenHave you ever used an automatic dishwasher? When a dishwasher door is opened a few minutes after the end of the washing cycle, you will find the ceramics and the heavy metal items will be completely dry. However, anything made of plastic will still be wet. This happens because plastic has a relatively low specific heat capacity, which means that it does not retain as much heat as the other material items and hence is not able to evaporate off the water droplets as quickly. In this article, we will learn all about specific heat capacity and investigate this property in different materials!

Specific heat capacity is a measure of how much energy is needed to raise the temperature of a material and is defined as follows:

The **specific heat capacity **of a substance is the energy required to raise the temperature of \( 1\,\mathrm{kg} \) of the substance by \( 1^\circ\mathrm C \).

Although you will have an intuitive understanding of temperature as how hot or cold something is, it can also be useful to know the actual definition.

The **temperature **of a substance is the average kinetic energy of the particles within it.

Energy is always needed to raise the temperature of a material. As energy is supplied, the internal energy of the particles in the material increases. Different states of matter react somewhat differently when they are heated:

- Heating a gas causes the particles to move around more quickly.
- Heating solids causes the particles to vibrate more.
- Heating liquids results in a combination of increased vibrations and faster movement of the particles.

When you use a bunsen burner to heat a beaker of water, the **thermal energy** of the flame is transferred to the particles in the water, which causes them to vibrate more and move faster. Therefore, the thermal energy is converted to kinetic energy.

The energy required to increase the temperature of a substance by a certain amount depends on two factors:

- The mass - the amount of a substance there is. The greater the mass, the more energy will be required to heat it up.
- The material - the temperature of different materials will increase by different amounts when energy is applied to them.

The amount that a material heats up when energy is applied to it depends on its specific heat capacity, \( c \). The greater a material's specific heat capacity, the more energy is required for its temperature to increase by a given amount. The specific heat capacities of various materials are shown in the table below.

Type of material | Material | Specific heat capacity (\( \mathrm J\,\mathrm{kg}^{-1}\,\mathrm K^{-1} \)) |

Metals | Lead | 130 |

Copper | 385 | |

Aluminium | 910 | |

Non-metals | Glass | 670 |

Ice | 2100 | |

Ethanol | 2500 | |

Water | 4200 | |

Air | 1000 |

The table shows that non-metals generally have a higher specific heat capacity than metals. Also, water has a very high specific heat capacity compared to other materials. It's value is \( 4200\,\mathrm J\,\mathrm{kg}^{-1}\,\mathrm K^{-1} \), meaning that \( 4200\,\mathrm J \) of energy is required to heat up \( 1 \,\mathrm kg \) of water by \( 1\,\mathrm K \). It takes a lot of energy to heat up water and, on the other hand, water takes a long time to cool down.

The high specific heat capacity of water has an interesting consequence for the world's climate. The material that makes up the Earth's land has a low specific heat capacity compared to water. This means that in summer the land warms and cools more quickly compared to the sea. In winter, the land cools faster than the sea does.

People living a long distance from the sea have extremely cold winters and very hot summers. Those living on the coast or near the sea do not experience the same extreme climates because the sea acts as a reservoir of heat in the winter and remains cooler in the summer!

Now that we have discussed what factors affect how the temperature of a substance changes, we can state the specific heat capacity formula. The change in energy, \( \Delta E \), required to produce a certain change in temperature, \( \Delta\theta \), in a material of mass \( m \) and specific heat capacity \( c \) is given by the equation

$\Delta $

which in words can be written as

$\text{change}$

Notice that this equation relates the *change *in energy to the *change *in temperature. The temperature of a substance decreases when energy is taken away from it, in which case the quantities \( \Delta E \) and \( \Delta\theta \) will be negative.

As you may have noticed from the table in the section above, the SI unit for specific heat capacity is \( \mathrm J\,\mathrm{kg}^{-1}\,\mathrm K^{-1} \). It can be derived from the specific heat capacity equation. Let us first rearrange the equation to find an expression for the specific heat capacity on its own:

$c$

The SI units for the quantities in the equation are as follows:

- Joules \( \mathrm J \), for energy.
- Kilograms \( \mathrm{kg} \), for mass.
- Kelvin \( \mathrm K \), for temperature.

We can plug the units into the equation for specific heat capacity to find the SI unit for \( c \):

$u$

As we are only dealing with a change in temperature - a difference between two temperatures rather than a single temperature - the units can be either Kelvin, \( \mathrm K \), or degrees Celsius, \( ^\circ \mathrm C \). The Kelvin and Celsius scales have the same divisions and only differ in their starting points - \( 1\,\mathrm K \) is equal to \( 1 ^\circ\mathrm C \).

A short experiment can be performed to find the specific heat capacity of a block of material, such as aluminium. Below is a list of the equipment and materials needed:

- Thermometer.
- Stopwatch.
- Immersion heater.
- Power supply.
- Ammeter.
- Voltmeter.
- Connecting wires.
- Aluminium block of known mass with holes for the thermometer and the immersion heater to be placed in.

This experiment uses an immersion heater to increase the temperature of an aluminium block so that the specific heat capacity of aluminium can be measured. The setup is shown in the image below. First, the immersion heater circuit needs to be constructed. The immersion heater should be connected to a power supply in series with an ammeter and placed in parallel with a voltmeter. Next, the heater can be placed inside the corresponding hole in the block and the same should be done for the thermometer.

Once everything is set up, turn on the power supply and start the stopwatch. Note the initial temperature of the thermometer. Take readings of the current from the ammeter and the voltage from the voltmeter every minute for a total of \( 10 \) minutes. When the time is up, note the final temperature.

In order to calculate the specific heat capacity, we must find the energy transferred to the block by the heater. We can use the equation

$E$

Once everything is set up, turn on the power supply and start the stopwatch. Note the initial temperature of the thermometer. Take readings of the current from the ammeter and the voltage from the voltmeter every minute for a total of \( 10 \) minutes. When the time is up, note the final temperature.

In order to calculate the specific heat capacity, we must find the energy transferred to the block by the heater. We can use the equation

$E$

where \( E \) is the energy transferred in Joules \( \mathrm J \), \( P \) is the power of the immersion heater in Watts \( \mathrm W \), and \( t \) is the heating time in seconds \( \mathrm s \). The power of the heater can be calculated by using

$P$

where \( I \) is the ammeter current in Amps \( \mathrm A \), and \( V \) is the voltage measured by the voltmeter in volts \( \mathrm V \). You should use your average current and voltage values in this equation. This means that the energy is given by

$E$

We already found an equation for specific heat capacity as

$c$

Now that we have an expression for the energy transferred to the aluminium block, we can substitute this into the specific heat capacity equation to get

$c$

After completing this experiment, you will have all of the quantities needed to calculate the specific heat capacity of aluminium. This experiment can be repeated to find the specific heat capacities of different materials.

There are several sources of error in this experiment that should be avoided or noted:

- The ammeter and voltmeter must both be initially set to zero so that the readings are correct.
- A small amount of energy is dissipated as heat in the wires.
- Some energy supplied by the immersion heater will be wasted - it will heat up the surroundings, the thermometer, and the block. This will result in the measured specific heat capacity being less than the true value. The proportion of wasted energy can be reduced by insulating the block.
- The thermometer must be read at eye level to record the correct temperature.

The equations discussed in this article can be used for many practice questions about specific heat capacity.

**Question**

An outdoor swimming pool needs to be heated up to a temperature of \( 25^\circ\mathrm C \). If its initial temperature is \( 16^\circ\mathrm C \) and the total mass of water in the pool is \( 400,000\,\mathrm kg \), how much energy is required to make the pool the correct temperature?

**Solution**

The specific heat capacity equation is

$\Delta $

We need the mass of water in the pool, the specific heat capacity of water and the change in temperature of the pool to compute the energy required to heat it up. The mass is given in the question as \( 400,000\,\mathrm kg \). The specific heat capacity of water was given in the table earlier in the article and is \( 4200\,\mathrm J\,\mathrm{kg}^{-1}\,\mathrm K^{-1} \). The change in temperature of the pool is the final temperature minus the initial temperature, which is

$\Delta $

All of these values can be plugged into the equation to find the energy as

$\u2206$

**Question**

An immersion heater is used to heat an aluminium block of mass \( 1\,\mathrm{kg} \), which has an initial temperature of \( 20^\circ\mathrm C \). If the heater transfers \( 10,000\,\mathrm J \) to the block, what final temperature does the block reach? The specific heat capacity of aluminium is \( 910\,\mathrm J\,\mathrm{kg}^{-1}\,\mathrm K^{-1} \).

**Solution**

For this question, we must once again use the specific heat capacity equation

$\Delta $

which can be rearranged to give an expression for the change in temperature, \( \Delta\theta \) as

$\Delta $

The change in energy is \( 10,000\,\mathrm J \), the mass of the aluminium block is \( 1\,\mathrm{kg} \) and the specific heat capacity of aluminium is \( 910\,\mathrm J\,\mathrm{kg}^{-1}\,\mathrm K^{-1} \). Substituting these quantities into the equation gives the change in temperature as

$\Delta $

The final temperature, \( \theta_{\mathrm F} \) is equal to the temperature change added to the initial temperature:

${\theta}_{\mathrm{F}}$

- The specific heat capacity
- The energy required to increase the temperature of a substance depends on its mass and the type of material.
- The greater a material's specific heat capacity, the more energy is required for its temperature to increase by a given amount.
- Metals generally have a higher specific heat capacity than non-metals.
- Water has a high specific heat capacity compared to other materials.
- The change in energy, \( \Delta E \), required to produce a certain change in temperature, \( \Delta\theta \), in a material of mass \( m \) and specific heat capacity \( c \) is given by the equation
\( \Delta E=mc\Delta\theta \).

The SI unit for specific heat capacity is \( \mathrm J\;\mathrm{kg}^{-1}\;\mathrm K^{-1} \).

Degrees Celsius can be exchanged for Kelvin in the units for specific heat capacity as \( 1^\circ \mathrm C \) is equal to \( 1\;\mathrm K \).

The specific heat capacity of a block of a certain material can be found by heating it with an immersion heater and using the equation \( E=IVt \) to find the energy transferred to the block from the heater's electrical circuit.

The symbol for specific heat capacity is *c* and its unit is J kg^{-1 }K^{-1}.

If you want to raise the temperature of an object by a certain amount, how does the energy required depend on the object's mass?

The energy required is directly proportional to the object's mass.

When attempting to measure specific heat capacity, the Kelvin temperature scale should always be used. Is this statement true or false?

False.

An electric heater is used to transfer the same amount of energy to a block of aluminium and a block of copper both with the same mass and initially at the same temperature. The copper block reaches a higher temperature. Which metal has the higher specific heat capacity?

Aluminium.

Water has a relatively high specific heat capacity compared to other materials. Is this statement true or false?

True.

What is the equation for the energy transferred by an electrical heater which has an average voltage across it of \( V \), an average current flowing through it of \( I \) and is switched on for \( t \) seconds?

\( E=IVt \).

Would it take longer to heat up a tank of water or a larger tank of air of the same *mass*?

A tank of water.

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