PARTICLE MODEL OF MATTER · CHAPTER 6 OF 10

Specific heat capacity

Heating a small amount of water takes less energy than heating a much larger amount through the same temperature rise. Different materials also need different amounts of energy. What determines how much energy a substance needs to warm up?

By the end, you’ll be able to…

  • Describe how energy input, mass and material affect temperature rise.
  • Define specific heat capacity and interpret its units.
  • Use ΔE = m × c × ΔT to calculate energy transfers and temperature changes.
  • Describe how to investigate specific heat capacity and explain the effect of energy losses.

What affects the temperature rise?

In Chapter 5, you saw that a temperature rise means an increase in the particles’ average kinetic energy. The size of that rise depends on the energy transferred into the substance, its mass and the material it is made from.

For the same mass and material, transferring more energy produces a larger temperature rise. For the same material and energy input, a larger mass has a smaller temperature rise. Different materials can also have different temperature rises for the same mass and energy input.

These comparisons assume that no change of state occurs and that the stated energy is transferred to the sample.

INTERACTIVE EXPLORER

Same heating, different result?

Choose a comparison. Predict which sample will have the greater temperature rise, then select Play heating.

Same material and mass. Different amounts of energy transferred.

Both samples contain 0.50kg of water and start at 20°C. Which will have the greater temperature rise?

Sample A · Water

Mass: 0.50kg
Energy transferred to the water 21,000J
Water temperature
20°C Warming… 30°C
Temperature rise: 0°C Temperature rise: 10°C

Sample B · Water

Mass: 0.50kg
Energy transferred to the water 63,000J
Water temperature
20°C Warming… 50°C
Temperature rise: 0°C Temperature rise: 30°C

Sample B receives three times as much energy and has three times the temperature rise: 30°C rather than 10°C.

For the same mass and material, more energy transferred produces a greater temperature rise.

Same material and energy transferred. Different masses.

Both samples start at 20°C and receive 63,000J of energy. Which will have the greater temperature rise?

Sample A · Water

Mass: 0.50kg
Energy transferred to the water 63,000J
Water temperature
20°C Warming… 50°C
Temperature rise: 0°C Temperature rise: 30°C

Sample B · Water

Mass: 1.0kg
Energy transferred to the water 63,000J
Water temperature
20°C Warming… 35°C
Temperature rise: 0°C Temperature rise: 15°C

Sample B has twice the mass but receives the same energy. Its temperature rise is half as large: 15°C rather than 30°C.

For the same material and energy input, a larger mass has a smaller temperature rise.

Same mass and energy transferred. Different materials.

Both samples have a mass of 1.0kg, start at 20°C and receive 36,000J of energy. Which will have the greater temperature rise?

Sample A · Aluminium

Mass: 1.0kg
c = 900 J/kg°C
Energy transferred to the aluminium 36,000J
Aluminium temperature
20°C Warming… 60°C
Temperature rise: 0°C Temperature rise: 40°C

Sample B · Water

Mass: 1.0kg
c = 4200 J/kg°C
Energy transferred to the water 36,000J
Water temperature
20°C Warming… 28.6°C
Temperature rise: 0°C Temperature rise: about 8.6°C

Aluminium warms by 40°C; water warms by about 8.6°C. Water has a higher specific heat capacity, so the same mass needs more energy for each degree of warming.

Equal masses receiving equal amounts of energy can have different temperature rises because they are different materials.

Energy values show the energy transferred into each sample during heating, not its total internal energy. Each sample already has internal energy at 20°C.

This idealised model ignores energy losses and energy absorbed by the apparatus, so energy transferred into the sample equals its increase in internal energy. No change of state occurs. Orange indicates active heating; it is not a real glow. The animation timescale and thermometer columns are illustrative.

What is specific heat capacity?

Specific heat capacity is the energy required to raise the temperature of 1kg of a substance by 1°C, without changing its state.

Its symbol is c and its unit is J/kg°C: joules per kilogram per degree Celsius.

For example, water has a specific heat capacity of approximately 4200 J/kg°C. This means 4200J is needed to warm 1kg of water by 1°C.

A higher specific heat capacity means more energy is needed for the same mass and temperature rise. If equal masses receive equal energy inputs, the material with the higher specific heat capacity warms less.

Specific heat capacity is a property of the material. Doubling the mass doubles the energy needed for the same temperature rise, but leaves its specific heat capacity unchanged.

Why does the sea warm and cool slowly?

Water has a high specific heat capacity, so a large amount of energy is needed to change its temperature. This helps explain why the sea warms more slowly than the beach and retains summer warmth into early autumn.

Early summer

Warming gradually

Cooler sea

A large mass of water warms gradually as it gains energy overall.

Early autumn

Cooling gradually

Still warm from summer

The sea retains warmth while gradually losing energy overall.

Water has a high specific heat capacity. A large mass of seawater needs a large energy transfer to warm up—and must transfer a large amount of energy out to cool down. It therefore warms and cools gradually, retaining energy absorbed during summer. This can keep the sea pleasantly warm for a beach holiday in early autumn.

* Energy enters and leaves the sea in both seasons. This is an illustrative seasonal comparison; sea temperatures vary with location, weather and currents. Sun rays and thermometer levels are illustrative. The thermometers show gradual changes, then reset.

Calculating energy transfers

The energy transferred to a substance during a temperature change is calculated using:

The equation

energy transferred = mass × specific heat capacity × temperature change

ΔE = m × c × ΔT

ΔE= change in thermal energy in joules (J)

m=mass in kilograms (kg)

c=specific heat capacity (J/kg°C)

ΔT=change in temperature (°C)

*This equation applies to a temperature change without a change of state. The energy needed for melting or boiling is explored in the next chapter.

Temperature change

The symbol Δ means “change in”. Temperature change can be written as ΔT or Δθ.

Temperature change = final temperature − initial temperature

Water warming from 20°C to 35°C has a temperature rise of:

35 − 20 = 15°C

Use the temperature rise of 15°C in the equation.

Mass in kilograms

Convert grams to kilograms before using the equation.

Mass in kg = mass in g ÷ 1000

For example:

500 ÷ 1000 = 0.50 kg

To calculate energy:

ΔE = m × c × ΔT

A kettle heats 0.50 kg of water from 20°C to 60°C. Water has a specific heat capacity of 4200 J/kg°C. Calculate the energy transferred to the water.

ΔT = 60 − 20 = 40°C
ΔE = m × c × ΔT
ΔE = 0.50 × 4200 × 40
ΔE = 84,000 J

To calculate mass:

m = ΔE ÷ (c × ΔT)

Water in a hot-water bottle receives 25,200 J of energy as its temperature rises by 10°C. Water has a specific heat capacity of 4200 J/kg°C. Calculate its mass.

m = ΔE ÷ (c × ΔT)
m = 25,200 ÷ (4200 × 10)
m = 25,200 ÷ 42,000
m = 0.60 kg

To calculate specific heat capacity:

c = ΔE ÷ (m × ΔT)

A 1.0 kg metal block receives 9000 J of energy and its temperature rises by 10°C. Calculate its specific heat capacity.

c = ΔE ÷ (m × ΔT)
c = 9000 ÷ (1.0 × 10)
c = 9000 ÷ 10
c = 900 J/kg°C

To calculate temperature change:

ΔT = ΔE ÷ (m × c)

A 2.0 kg aluminium pan receives 9000 J of energy. Aluminium has a specific heat capacity of 900 J/kg°C. Calculate the pan’s temperature rise.

ΔT = ΔE ÷ (m × c)
ΔT = 9000 ÷ (2.0 × 900)
ΔT = 9000 ÷ 1800
ΔT = 5°C

In these examples, the stated energy is transferred to the substance being heated. Energy losses to the surroundings are not included

Measuring specific heat capacity

You need to measure the sample’s mass, its temperature rise and the energy supplied during heating.

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PRACTICAL FOCUS · MEASURING SPECIFIC HEAT CAPACITY

What do you need to measure?

Measure the water’s mass, the energy supplied during heating and its temperature rise.

Apparatus for measuring the specific heat capacity of water A low-voltage supply connects through an electrical energy meter to an immersion heater in an insulated water container. A thermometer measures the water temperature. A lid reduces energy transfer to the surroundings. The water mass is measured separately using a balance. Low-voltage supply Energy meter joules (J) IN OUT Thermometer Lid Heater Water Insulated container Electrical leads shown schematically

The heater supplies energy; the thermometer measures temperature. The lid and insulation reduce energy transfer to the surroundings.

  1. 1

    Measure the mass

    Tare the balance with the empty container, then add water. Record its mass in kilograms.

  2. 2

    Record the starting readings

    Record the initial water temperature and energy-meter reading.

  3. 3

    Heat and measure

    Switch on the heater and stir gently. Switch it off, then record the final temperature and energy-meter reading.

  4. 4

    Calculate the changes

    Subtract initial from final readings to find the energy supplied and temperature rise.

  5. 5

    Calculate specific heat capacity

    c = ΔE ÷ (m × ΔT)

The energy meter measures energy supplied to the heater. Using this as the water’s energy gain assumes negligible energy transfer to the apparatus and surroundings. The next section explains the effect of these losses.

What about energy losses?

The energy meter measures electrical energy supplied to the heater. Some of this warms the heater and container, and some is transferred to the surroundings. It does not all warm the sample.

If you use the full meter reading as the sample’s energy gain, you will usually calculate a specific heat capacity that is too high. The measured temperature rise is smaller than it would be if all that energy warmed the sample.

 

Insulation and a lid reduce energy transfer to the surroundings. Good thermal contact helps in a metal-block investigation; gentle stirring helps make water-temperature readings representative of the whole sample.

Repeating measurements helps assess consistency, but does not remove a systematic error caused by energy losses.

Alternatively, electrical energy supplied can be calculated using E = P × t, or E = V × I × t when voltage and current are measured and remain approximately constant. Time must be in seconds.

* Energy supplied during heating warms the water and apparatus, while some transfers to the surroundings.

INTERACTIVE QUICK CHECK
Choose one answer for each question. You can change your answers.
1 What does a specific heat capacity of 900 J/kg°C mean?
2 The same energy is transferred to two samples of water. Their masses are 0.50kg and 1.0kg. Which has the greater temperature rise? Assume no energy losses or change of state.
3 A 0.50kg sample has a specific heat capacity of 800 J/kg°C. Its temperature rises from 20°C to 30°C. How much energy is transferred to the sample?
INTERACTIVE CHALLENGE

A student calculates the specific heat capacity of water using all the electrical energy supplied by a heater. Some energy warmed the container and surroundings. Will the calculated value be too high or too low? Explain your answer.

Use c = ΔE ÷ (m × ΔT) to support your reasoning.

The calculated value will be too high.

Not all the electrical energy supplied increases the water’s internal energy. Some warms the apparatus and some transfers to the surroundings during heating.

In c = ΔE ÷ (m × ΔT), the student therefore uses an energy value that is too large for the measured mass and temperature rise, giving a value of c that is too high.

Insulation and a lid reduce energy transfer to the surroundings.

CHAPTER 6

Chapter summary

  • Specific heat capacity is the energy needed to raise the temperature
    of 1kg of a substance by 1°C. Its unit is J/kg°C.
  • Use ΔE = m × c × ΔT for a temperature change without a change of state.
    Use mass in kilograms and energy in joules.
  • Temperature change is final temperature minus initial temperature.
    It may be written as ΔT or Δθ.
  • For the same mass and material, transferring more energy produces
    a greater temperature rise.
  • For the same material and energy input, a larger mass has a smaller
    temperature rise.
  • A higher specific heat capacity means more energy is needed for
    the same mass and temperature rise.
  • The same equation applies to cooling: use the temperature fall
    to calculate the amount of energy transferred out.
  • Measure mass, energy supplied and temperature rise to investigate
    specific heat capacity.
  • Some supplied energy warms the apparatus or transfers to the
    surroundings during heating. Using all the supplied energy as
    the sample’s energy gain can make the calculated specific heat
    capacity too high.

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