Chapter 6 of 8

All 8 chapters
  1. 1Calculating energy stores, readunderstood
  2. 2Springs and elastic energy, readunderstood
  3. 3Work done, readunderstood
  4. 4Power, readunderstood
  5. 5Efficiency, readunderstood
  6. 6Specific heat capacity, readunderstood
  7. 7Energy resources, readunderstood
  8. 8Energy at home, readunderstood

S4.9.6 Chapter 6

Specific heat capacity

The same energy is given to 1 kg of water and to 1 kg of aluminium. Which one gets hotter?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

  • You picked D. Read on. At the end, the chapter comes back to your guess.

By the end you can

  • Use specific heat capacity to calculate the energy for a temperature rise
  • Explain why coasts have milder weather than places far inland

Put a 200 g metal pan and 200 g of water in the same sunshine. The pan is hot to touch long before the water warms. The masses are the same, so the difference comes from the materials. Different materials need different amounts of energy to warm up.

The specific heat capacity of a material is the energy needed to warm 1 kg of it by 1 °C. It is written c. Its unit is joules for each kg for each °C, or J/kg °C.

Water has a specific heat capacity of 4 200 J/kg °C. So it takes 4 200 J to warm 1 kg of water by 1 °C. A high number means a lot of energy for each degree. So the temperature rises slowly.

A table of specific heat capacities with the columns Material and c (J/kg °C). Water 4 200. Aluminium 900. Iron 450. Copper 390. Lead 130.Materialc (J/kg °C)water4 200aluminium900iron450copper390lead130
Approximate values. Water is far higher than the metals.

The energy needed depends on three things. The mass m in kg, the specific heat capacity c, and the change in temperature ΔT in °C. Here ΔT is the final temperature minus the start temperature.

change in thermal energy = mass × specific heat capacity × change in temperature. In symbols, ΔE = m × c × ΔT.

Worked example

How much energy is needed to warm 2 kg of water from 20 °C to 70 °C? The specific heat capacity of water is 4 200 J/kg °C.

  1. ΔE = m × c × ΔT.
  2. ΔT = 70 − 20 = 50 °C.
  3. ΔE = 2 × 4 200 × 50 = 420 000 J = 420 kJ.

Answer: 420 kJ

Mind the units. The mass must be in kg. Divide grams by 1 000. And ΔT is the change, not the final temperature. A rise from 20 °C to 70 °C is a change of 50 °C, not 70 °C.

Worked example

A block of aluminium has a mass of 500 g. It is heated from 15 °C to 55 °C. Its specific heat capacity is 900 J/kg °C. How much energy does it take in?

  1. m = 500 g = 0.5 kg.
  2. ΔT = 55 − 15 = 40 °C.
  3. ΔE = 0.5 × 900 × 40 = 18 000 J = 18 kJ.

Answer: 18 kJ

The formula can be turned around. c = ΔE ÷ (m × ΔT) gives the specific heat capacity. And ΔT = ΔE ÷ (m × c) gives the temperature rise. A heater of power P running for a time t supplies ΔE = P × t.

Two containers side by side, each holding water 0.5 kg with a heater and a thermometer. The left one reads 20 °C and the right one reads 30 °C.20 °Cheaterwater 0.5 kg30 °Cheaterwater 0.5 kg
A 100 W heater in 0.5 kg of water, before and after 210 s.

Worked example

A 100 W heater warms 0.5 kg of water for 210 s. All its energy goes into the water. By how many °C does the temperature rise?

  1. The energy supplied is ΔE = P × t = 100 × 210 = 21 000 J.
  2. ΔE = m × c × ΔT, so ΔT = ΔE ÷ (m × c).
  3. m × c = 0.5 × 4 200 = 2 100 J/°C.
  4. ΔT = 21 000 ÷ 2 100 = 10 °C.

Answer: 10 °C

This test is done in a laboratory with a low voltage heater. It is not done at home, because mains electricity and water are dangerous together. In a real test some heat escapes to the air, so the water warms a little less than the sum says.

Now to the coast. Water has a specific heat capacity of 4 200 J/kg °C. Dry soil and rock have about 800 J/kg °C. So the same sunshine warms the land about five times as much as the same mass of sea water.

At night the same thing happens in reverse. The land loses heat and cools fast. The sea needs to lose a lot of energy to cool by 1 °C, so it cools slowly. The sea also mixes, which spreads its heat through deep water.

A graph of temperature in degrees Celsius against the hour of the day. The axes are labelled hour of the day (h), from 0 to 24 in steps of 4, and temperature (°C), from 12 to 32 in steps of 4. A curve labelled land swings between 14 and 30. A curve labelled sea stays close to 20.landsea04812162024121620242832hour of the day (h)temperature (°C)
A made-up day to show the pattern. The temperature scale starts at 12 °C, not 0. The land swings from 14 °C to 30 °C. The sea only moves between 19 °C and 21 °C.

Winds that blow from the sea carry this steadiness onto the coast. So a coast is usually cooler than inland when it is hot, and warmer than inland when it is cold. Far inland there is no sea to hold the temperature steady, so the range is bigger.

This holds where the wind comes from the sea and where no cold sea current runs along the coast. A coast can be different in a place where the wind blows from the land.

A wrong idea many people have

A material with a high specific heat capacity heats up faster.

Why it feels right: A high number sounds like a quick or strong thing.

The test that breaks it: Give 4 200 J to 1 kg of water: it warms by 1 °C. Give 4 200 J to 1 kg of lead, with c = 130: it warms by 4 200 ÷ 130, about 32 °C.

What is true: A high specific heat capacity means a lot of energy is needed for each degree. So the material heats up slowly, and cools slowly too.

Try it

Soil and water in the sun

You need: two identical dark cups or tins, dry soil, cold water, a sunny spot, a thermometer, if you have one.

Be safe: Never look at the Sun. Touch only with a finger. Do not use hot water, and do not leave the water where a child or an animal may drink it.

  1. Fill one cup to the top with dry soil. Fill the other to the top with water. Use the same cup size.
  2. Put both cups in the same sunny place. Do not move them.
  3. After an hour, feel the top of the soil and the water with a finger. If you have a thermometer, measure both.
  4. Move the cups into the shade in the evening and feel them again an hour later.
  5. Write down which cup warmed more and which cooled faster.

What you will see: The soil feels warmer after the hour in the sun. Late in the day it also cools faster. The soil heats up and cools down more quickly than the water because it has a lower specific heat capacity. This is a rough test, because the Sun also warms the soil and the water in different ways.

If you cannot try it, this is what happens.
A table of example temperatures with the columns Cup, Start and After sun. Soil: 22 °C at the start and 38 °C after sun. Water: 22 °C at the start and 29 °C after sun.CupStartAfter sunsoil22 °C38 °Cwater22 °C29 °C

Key words

specific heat capacity
the energy needed to raise the temperature of 1 kg of a material by 1 °C, in J/kg °C.
change in temperature
the final temperature minus the start temperature, written ΔT.

Back to your guess

The answer is C: The aluminium, because it takes less energy to warm it each degree.

Aluminium has a much lower specific heat capacity than water. The same energy warms 1 kg of aluminium by a much larger number of degrees than 1 kg of water.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. Which needs more energy to warm by 1 °C, 1 kg of water or 1 kg of copper? Look at the table of values.

    Tap your answer.

    1. Not this one. Density is not the reason. What matters is the specific heat capacity, and copper is far lower than water. Try another.

    2. Not this one. The mass is the same, but the materials differ. Water needs much more energy for each degree. Try another.

    3. Not this one. A lower specific heat capacity means less energy for each degree, so copper needs less, not more. Try another.

    4. Yes, that is right. Water needs about 4 200 J for each kg and each °C, copper only about 390 J. So water needs about ten times as much energy.

    See the answer

    Water, because its specific heat capacity is higher.

    Water needs about 4 200 J for each kg and each °C, copper only about 390 J. So water needs about ten times as much energy.

    • If you chose A: Density is not the reason. What matters is the specific heat capacity, and copper is far lower than water.
    • If you chose B: The mass is the same, but the materials differ. Water needs much more energy for each degree.
    • If you chose C: A lower specific heat capacity means less energy for each degree, so copper needs less, not more.

  2. How much energy raises the temperature of 3 kg of water by 5 °C? The specific heat capacity of water is 4 200 J/kg °C.

    Two containers side by side, each holding water 3 kg with a heater and a thermometer. The left one reads 20 °C and the right one reads 25 °C.20 °Cheaterwater 3 kg25 °Cheaterwater 3 kg
    Tap your answer.

    1. Yes, that is right. ΔE = m × c × ΔT = 3 × 4 200 × 5 = 63 000 J = 63 kJ.

    2. Not this one. You left out the mass. The energy is m × c × ΔT, so 3 kg counts. Try another.

    3. Not this one. You left out the change in temperature. The 5 °C must be multiplied in. Try another.

    4. Not this one. You divided by the change in temperature. All three numbers are multiplied. Try another.

    See the answer

    63 kJ

    ΔE = m × c × ΔT = 3 × 4 200 × 5 = 63 000 J = 63 kJ.

    • If you chose B: You left out the mass. The energy is m × c × ΔT, so 3 kg counts.
    • If you chose C: You left out the change in temperature. The 5 °C must be multiplied in.
    • If you chose D: You divided by the change in temperature. All three numbers are multiplied.

  3. A piece of iron of mass 250 g is heated by 20 °C. The specific heat capacity of iron is 450 J/kg °C. How much energy does it take in?

    Tap your answer.

    1. Not this one. You left out the mass. The energy is m × c × ΔT, so the 250 g counts. Try another.

    2. Yes, that is right. m = 250 g = 0.25 kg. ΔE = m × c × ΔT = 0.25 × 450 × 20 = 2 250 J = 2.25 kJ.

    3. Not this one. You left out the change in temperature. The 20 °C must be multiplied in. Try another.

    4. Not this one. You used 250 as the mass, but the mass must be in kg. 250 g is 0.25 kg. Go to M2.6.4. Try another.

    See the answer

    2.25 kJ

    m = 250 g = 0.25 kg. ΔE = m × c × ΔT = 0.25 × 450 × 20 = 2 250 J = 2.25 kJ.

    • If you chose A: You left out the mass. The energy is m × c × ΔT, so the 250 g counts.
    • If you chose C: You left out the change in temperature. The 20 °C must be multiplied in.
    • If you chose D: You used 250 as the mass, but the mass must be in kg. 250 g is 0.25 kg. Go to M2.6.4.

  4. Why are coastal places usually milder than places far inland?

    Tap your answer.

    1. Yes, that is right. Water needs a lot of energy for each degree, so the sea changes temperature slowly. Winds from the sea carry that steadiness to the coast.

    2. Not this one. The sea can be cooler than the land in summer. It is the slow change of temperature that matters. Try another.

    3. Not this one. The Sun shines on the coast and the inland in the same way. The difference comes from the sea. Try another.

    4. Not this one. The sea absorbs plenty of energy. It just takes a lot of energy to warm it by one degree. Try another.

    See the answer

    The sea has a high specific heat capacity, so it warms and cools slowly and keeps the air steady.

    Water needs a lot of energy for each degree, so the sea changes temperature slowly. Winds from the sea carry that steadiness to the coast.

    • If you chose B: The sea can be cooler than the land in summer. It is the slow change of temperature that matters.
    • If you chose C: The Sun shines on the coast and the inland in the same way. The difference comes from the sea.
    • If you chose D: The sea absorbs plenty of energy. It just takes a lot of energy to warm it by one degree.

  5. A 4 kg block takes in 3 600 J of energy. Its temperature rises by 3 °C. What is its specific heat capacity?

    Tap your answer.

    1. Not this one. You left out the mass. c = ΔE ÷ (m × ΔT), so divide by the 4 kg too. Try another.

    2. Not this one. You left out the change in temperature. Divide by the 3 °C as well as the mass. Try another.

    3. Not this one. You multiplied all three numbers. For c you divide the energy by the mass and the temperature change. Try another.

    4. Yes, that is right. c = ΔE ÷ (m × ΔT) = 3 600 ÷ (4 × 3) = 3 600 ÷ 12 = 300 J/kg °C.

    See the answer

    300 J/kg °C

    c = ΔE ÷ (m × ΔT) = 3 600 ÷ (4 × 3) = 3 600 ÷ 12 = 300 J/kg °C.

    • If you chose A: You left out the mass. c = ΔE ÷ (m × ΔT), so divide by the 4 kg too.
    • If you chose B: You left out the change in temperature. Divide by the 3 °C as well as the mass.
    • If you chose C: You multiplied all three numbers. For c you divide the energy by the mass and the temperature change.

  6. A 2 kg lump of iron is heated from 20 °C to 45 °C. The specific heat capacity of iron is 450 J/kg °C. How much energy does it take in?

    Two containers side by side, each holding iron 2 kg with a heater and a thermometer. The left one reads 20 °C and the right one reads 45 °C.20 °Cheateriron 2 kg45 °Cheateriron 2 kg
    Tap your answer.

    1. Not this one. You used the final temperature, 45 °C. The sum needs the change, 45 − 20 = 25 °C. Try another.

    2. Not this one. You added the two temperatures. The sum needs the difference, 45 − 20 = 25 °C. Try another.

    3. Yes, that is right. ΔT = 45 − 20 = 25 °C. ΔE = m × c × ΔT = 2 × 450 × 25 = 22 500 J = 22.5 kJ.

    4. Not this one. You used the start temperature, 20 °C. The sum needs the change, 45 − 20 = 25 °C. Try another.

    See the answer

    22.5 kJ

    ΔT = 45 − 20 = 25 °C. ΔE = m × c × ΔT = 2 × 450 × 25 = 22 500 J = 22.5 kJ.

    • If you chose A: You used the final temperature, 45 °C. The sum needs the change, 45 − 20 = 25 °C.
    • If you chose B: You added the two temperatures. The sum needs the difference, 45 − 20 = 25 °C.
    • If you chose D: You used the start temperature, 20 °C. The sum needs the change, 45 − 20 = 25 °C.

  7. From an earlier chapter

    A kettle takes in 2 000 J of electrical energy and 1 800 J of it heats the water. What is its efficiency?

    Tap your answer.

    1. Yes, that is right. efficiency = useful output ÷ total input = 1 800 ÷ 2 000 = 0.9, which is 90%.

    2. Not this one. You used the wasted energy, 200 J. Use the useful energy that heats the water. Go to S4.9.5. Try another.

    3. Not this one. You divided the input by the output, which is upside down. An efficiency is never above 100%. Go to S4.9.5. Try another.

    See the answer

    90%

    efficiency = useful output ÷ total input = 1 800 ÷ 2 000 = 0.9, which is 90%.

    • If you chose B: You used the wasted energy, 200 J. Use the useful energy that heats the water. Go to S4.9.5.
    • If you chose C: You divided the input by the output, which is upside down. An efficiency is never above 100%. Go to S4.9.5.

  8. A 1.5 kg mass of water is heated from 20 °C to 60 °C. The specific heat capacity of water is 4 200 J/kg °C. How much energy does it take in, in kJ?

    See the answer

    252 kJ

    ΔT = 60 − 20 = 40 °C. ΔE = 1.5 × 4 200 × 40 = 252 000 J = 252 kJ.

  9. A 200 g piece of copper is heated by 30 °C. The specific heat capacity of copper is 390 J/kg °C. How much energy does it take in, in joules?

    See the answer

    2 340 J

    m = 200 g = 0.2 kg. ΔE = 0.2 × 390 × 30 = 2 340 J.

  10. A 200 W heater warms 2 kg of water for 105 s. All its energy goes into the water. By how many °C does the temperature rise?

    See the answer

    2.5 °C

    ΔE = P × t = 200 × 105 = 21 000 J. Then ΔT = ΔE ÷ (m × c). So ΔT = 21 000 ÷ (2 × 4 200) = 21 000 ÷ 8 400 = 2.5 °C.

Keep it

Copy this into your exercise book.

ΔE = m × c × ΔT. Mass in kg. ΔT is the change. Water: c = 4 200 J/kg °C. High c: slow to warm and slow to cool. The sea keeps coasts mild.

Draw: A container of water with a heater and a thermometer, and ΔE = m × c × ΔT beside it.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

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