Specific heat capacity
By the end of this chapter you will be able to
- Calculate the energy needed to warm a mass of a material by a given temperature rise
- Rearrange E = mc(change in temperature) to find any one of the quantities
- Explain what specific heat capacity means, and why water is used in heating systems
- Describe how to measure the specific heat capacity of a metal, and explain how heat loss affects the result
You should already know
- Energy stores and transfers (Chapter 1)
- Efficiency and power calculations (Chapter 2)
Where this leads
- Particle model and internal energy (Chapter 9)
- Energy resources and the energy supply (Chapter 4)
GCSE Physics. Content follows the GCSE (9-1) subject content for England (Edexcel and AQA). Content references: Use E = m c change in theta to calculate the energy needed to change the temperature of a material, Explain specific heat capacity as the energy needed per kilogram per degree Celsius, Rearrange the equation to find the mass, the temperature change or the specific heat capacity, Describe the practical to determine the specific heat capacity of a metal block using an electric heater, a thermometer and insulation, Evaluate the effect of energy lost to the surroundings on a measured value.
Exercise P Do you remember?
Method. Multiply the two numbers.
- Multiply. Twice 4.5 is 9.2 x 4.5 = 9
- Write the answer. 9
Check. 9 / 2 = 4.5. ✓
Method. Multiply the two numbers.
- Multiply. 3 x 3 = 9, then add the two zeros: 9000.30 x 300 = 9000
- Write the answer. 9000
Check. 9000 / 300 = 30. ✓
Method. Divide the energy by the time.
- Divide. P = 2000 / 100 = 20 W.2000 / 100 = 20
- Write the answer. 20
Check. 20 W x 100 s = 2000 J. ✓
Method. One kilogram is 1000 grams.
- Multiply. 0.5 x 1000 = 500 g.0.5 x 1000 = 500
- Write the answer. 500
Check. 500 g / 1000 = 0.5 kg. ✓
Method. Think about what insulation is for.
- Recall. An insulator slows the transfer of thermal energy, so it keeps heat in.slows heat
- Write the answer. insulator
Check. Insulation is an insulator. ✓
What is specific heat capacity?
The specific heat capacity, written c, of a material is the energy needed to raise the temperature of 1 kg of it by 1 °C. Its unit is joules per kilogram per degree Celsius, J/kg °C. A material with a high specific heat capacity needs a lot of energy to warm up, and it also gives out a lot of energy when it cools down.
| Substance | Specific heat capacity (J/kg degC) |
|---|---|
| water | 4200 |
| aluminium | 900 |
| iron | 450 |
| copper | 385 |
Key ideas
- Specific heat capacity c = energy needed to warm 1 kg by 1 °C.
- Its unit is J/kg °C.
- Water has a high specific heat capacity: 4200 J/kg °C.
- Metals have lower values, so they warm up quickly.
Method. Read the value from the table.
- Read. The specific heat capacity of iron is 450 J/kg °C.c = 450
- Interpret. So 450 J of energy warms 1 kg of iron by 1 °C.450 J for 1 kg, 1 °C
- Write the answer. 450 J
Check. Iron needs less energy than water for the same mass and rise. ✓
Method. Compare the two values.
- Water. Water has c = 4200 J/kg °C.4200
- Copper. Copper has c = 385 J/kg °C.385
- Compare. 4200 is much bigger than 385, so water needs more energy.4200 > 385
- Write the answer. water
Check. Water warms up slowly and cools slowly too. ✓
Method. Think about the specific heat capacity of water.
- Think. Water has a high specific heat capacity.high c
- Link. So a given mass of water can store a lot of energy for each degree it is warmed.stores energy
- Write the answer. Accept any sentence that says water has a high specific heat capacity, so it can store a lot of energy.
Check. Water is a useful store because it holds a lot of energy. ✓
Watch out
Watch out: specific heat capacity is for 1 kg and 1 °C. Do not confuse it with the energy you transfer in total; that depends on the mass and the rise.
Exam tip
Exam tip. A high specific heat capacity means “hard to heat up and hard to cool down”. Use that idea in explanations.
Exercise 3A Meaning of specific heat capacity
Method. Specific heat capacity is energy per kilogram per degree.
- Recall. It is joules per kilogram per degree Celsius.J/kg °C
- Write the answer. J/kg °C
Check. Energy (J) divided by mass (kg) and temperature rise (°C). ✓
Method. Read the table.
- Read. The specific heat capacity of water is 4200 J/kg °C.4200
- Write the answer. 4200
Check. ✓
Method. Look at the four values.
- Compare. Copper has 385, the lowest value in the table.385
- Write the answer. copper
Check. Copper warms up very quickly for its mass. ✓
Method. Find the largest value.
- Compare. Water has the largest value, 4200.4200 is the largest
- Write the answer. water
Check. Water has the highest specific heat capacity in this table. ✓
Method. Read the value for aluminium.
- Read. c for aluminium = 900 J/kg °C.900
- Write the answer. 900 J
Check. 900 J per kilogram per degree. ✓
Method. Compare each value with 4200.
- Aluminium. 900 is less than 4200.900 < 4200
- Iron. 450 is less than 4200.450 < 4200
- Copper. 385 is less than 4200.385 < 4200
- Write the answer. aluminium, iron, copper
Check. Only water has the value 4200 in the table. ✓
Method. Think about what the specific heat capacity measures.
- Think. It is the energy needed to warm 1 kg by 1 °C.energy needed
- Link. A high value means more energy is needed, so it warms more slowly.more slowly
- Write the answer. Accept any sentence that says a high value means it takes a lot of energy to warm up, so it heats more slowly.
Check. Copper warms more quickly than water for the same energy input. ✓
Method. Use the idea of specific heat capacity.
- Compare. Water has a higher specific heat capacity, so for the same energy it warms less than iron.water warms less
- Write the answer. yes
Check. Water needs more energy per degree than iron. ✓
Method. Think about the properties of the metal and the plastic.
- Metal. The metal has a low specific heat capacity and conducts heat well.low c, conducts
- Plastic. Plastic is an insulator and does not warm the hand.insulator
- Write the answer. Accept any answer that says metal warms up fast and conducts energy well, while a plastic handle stays cool to touch.
Check. ✓
Method. Use the value for water.
- Value. Water has c = 4200 J/kg °C.4200
- Explain. A lot of energy is needed to warm the water, so it warms slowly.warms slowly
- Write the answer. Accept any answer that says water has a high specific heat capacity, so it needs a lot of energy to warm up.
Check. ✓
Method. Multiply the specific heat capacity by the mass and the rise.
- Multiply. E = 900 x 3 x 1 = 2700 J.900 x 3 x 1 = 2700
- Write the answer. 2700 J
Check. 2700 / 3 = 900 J per kg for 1 °C. ✓
Method. The block with the lower specific heat capacity warms more for the same energy.
- Compare. X has the lower value, so it warms by more for the same energy.450 < 900
- Write the answer. X
Check. The temperature rise is twice as big for X. ✓
Method. Use the idea of specific heat capacity.
- Oil. Cooking oil has a lower specific heat capacity than water.lower c
- Link. For the same energy input, a lower c gives a bigger temperature rise.bigger rise
- Write the answer. Accept any answer that uses specific heat capacity: the oil has a lower value, so it warms by more for the same energy.
Check. ✓
Calculating the energy to warm a material
The energy needed to change the temperature of a material is:
E = m x c x Δθ
where E is the energy in joules, m is the mass in kilograms, c is the specific heat capacity in J/kg °C, and Δθ is the temperature change in °C. The more mass you have, and the bigger the rise, the more energy you need.
| Substance | Specific heat capacity (J/kg degC) |
|---|---|
| water | 4200 |
| aluminium | 900 |
| iron | 450 |
| copper | 385 |
Key ideas
- E = m x c x Δθ, in joules, with m in kilograms and Δθ in °C.
- A rise of 10 °C and a fall of 10 °C use the same numbers: use the size of the change.
- Convert grams to kilograms first: 500 g = 0.5 kg.
- Always write the unit J at the end of your answer.
Method. Use E = m x c x Δθ with c = 4200 J/kg °C.
- Take the numbers. m = 2 kg, c = 4200 J/kg °C, Δθ = 10 °C.m = 2, c = 4200, Δθ = 10
- Multiply. E = 2 x 4200 x 10 = 84000 J.2 x 4200 x 10 = 84000
- Write the answer. 84000 J
Check. 84000 / (4200 x 10) = 2 kg. ✓
Method. Find the temperature change first: 40 - 20 = 20 °C.
- Temperature change. Δθ = 40 - 20 = 20 °C.20
- Multiply. E = 0.5 x 900 x 20 = 9000 J.0.5 x 900 x 20 = 9000
- Write the answer. 9000 J
Check. 9000 / (900 x 20) = 0.5 kg. ✓
Method. Use E = m x c x Δθ.
- Take the numbers. m = 2 kg, c = 385 J/kg °C, Δθ = 5 °C.m = 2, c = 385, Δθ = 5
- Multiply. E = 2 x 385 x 5 = 3850 J.2 x 385 x 5 = 3850
- Write the answer. 3850 J
Check. 3850 / (2 x 5) = 385 J/kg °C. ✓
Watch out
Watch out: convert grams to kilograms before you use E = mcΔθ. 250 g is 0.25 kg, not 250 kg.
Exam tip
Exam tip. Write the equation, then the numbers with their units, then calculate. Give your answer in joules.
Exercise 3B Energy to change the temperature
Method. Use E = m x c x Δθ.
- Multiply. E = 1 x 4200 x 10 = 42000 J.1 x 4200 x 10 = 42000
- Write the answer. 42000 J
Check. 42000 / 10 = 4200 J/kg °C. ✓
Method. Use E = m x c x Δθ.
- Multiply. E = 2 x 450 x 10 = 9000 J.2 x 450 x 10 = 9000
- Write the answer. 9000 J
Check. 9000 / (2 x 10) = 450. ✓
Method. Use E = m x c x Δθ.
- Multiply. E = 3 x 4200 x 5 = 63000 J.3 x 4200 x 5 = 63000
- Write the answer. 63000 J
Check. 63000 / (3 x 5) = 4200. ✓
Method. Convert the mass if needed (it is already in kilograms).
- Multiply. E = 0.5 x 4200 x 20 = 42000 J.0.5 x 4200 x 20 = 42000
- Write the answer. 42000 J
Check. 42000 / (0.5 x 20) = 4200. ✓
Method. Convert 500 g to 0.5 kg first.
- Convert. 500 g = 0.5 kg.0.5 kg
- Multiply. E = 0.5 x 385 x 10 = 1925 J.0.5 x 385 x 10 = 1925
- Write the answer. 1925 J
Check. 1925 / (0.5 x 10) = 385. ✓
Method. The temperature change is 25 - 15 = 10 °C.
- Change. Δθ = 25 - 15 = 10 °C.10
- Multiply. E = 4 x 900 x 10 = 36000 J.4 x 900 x 10 = 36000
- Write the answer. 36000 J
Check. 36000 / (4 x 10) = 900. ✓
Method. Find the rise, then the energy.
- Rise. Δθ = 52 - 12 = 40 °C.40
- Multiply. E = 2 x 4200 x 40 = 336000 J.2 x 4200 x 40 = 336000
- Write the answer. 336000 J
Check. 336000 / (2 x 40) = 4200. ✓
Method. Find the temperature rise, then use the equation.
- Rise. Δθ = 100 - 18 = 82 °C.82
- Multiply. E = 1.5 x 4200 x 82 = 516600 J.1.5 x 4200 x 82 = 516600
- Write the answer. 516600 J
Check. 516600 / (1.5 x 4200) = 82. ✓
Method. Convert 200 g to kilograms.
- Convert. 200 g = 0.2 kg.0.2 kg
- Multiply. E = 0.2 x 4200 x 25 = 21000 J.0.2 x 4200 x 25 = 21000
- Write the answer. 21000 J
Check. 21000 / (0.2 x 25) = 4200. ✓
Method. Find the energy for each pan and subtract.
- Pan one. E = 1 x 4200 x 10 = 42000 J.42000 J
- Pan two. E = 1 x 4200 x 20 = 84000 J.84000 J
- Difference. 84000 - 42000 = 42000 J.84000 - 42000 = 42000
- Write the answer. 42000 J
Check. Doubling the rise doubles the energy. ✓
Method. Rearrange E = m c Δθ to c = E / (m Δθ).
- Rearrange. c = 4500 / (2 x 5) = 4500 / 10.4500 / 10
- Divide. c = 450 J/kg °C.4500 / 10 = 450
- Write the answer. 450 J/kg °C
Check. 2 x 450 x 5 = 4500 J. ✓
Method. Check the units.
- Units. kg x (J/kg °C) x °C.kg x J/kg °C x °C
- Cancel. The kg and °C cancel, leaving J.J
- Write the answer. Accept any answer that multiplies kg by J/kg °C by °C and gives J.
Check. ✓
Method. Check the energy from the heater, and the energy the water needs.
- Water. The water needs E = 1 x 4200 x 10 = 42000 J.42000 J
- Heater. A 4200 W heater for 60 s gives 4200 x 60 = 252000 J, which is not the same.252000 J
- Decide. The 4200 J is not the energy the water needs, so the pupil is not right.no
- Write the answer. no
Check. ✓
Rearranging the equation
E = m c Δθ can be rearranged to find any one of the four quantities:
m = E / (c x Δθ), c = E / (m x Δθ), Δθ = E / (m x c)
Cover the quantity you want with your finger, and the rest of the equation shows you what to do.
Key ideas
- To find Δθ: divide E by (m x c).
- To find m: divide E by (c x Δθ).
- To find c: divide E by (m x Δθ).
- Check your answer by putting it back into E = m c Δθ.
Method. Rearrange to Δθ = E / (m x c).
- Rearrange. Δθ = 84000 / (2 x 4200).84000 / 8400
- Divide. Δθ = 84000 / 8400 = 10 °C.84000 / 8400 = 10
- Write the answer. 10 °C
Check. 2 x 4200 x 10 = 84000 J. ✓
Method. Rearrange to m = E / (c x Δθ).
- Rearrange. m = 9000 / (900 x 10).9000 / 9000
- Divide. m = 9000 / 9000 = 1 kg.9000 / 9000 = 1
- Write the answer. 1 kg
Check. 1 x 900 x 10 = 9000 J. ✓
Method. Rearrange to c = E / (m x Δθ).
- Rearrange. c = 9000 / (0.5 x 40) = 9000 / 20.9000 / 20
- Divide. c = 9000 / 20 = 450 J/kg °C.9000 / 20 = 450
- Write the answer. 450 J/kg °C
Check. 0.5 x 450 x 40 = 9000 J. ✓
Watch out
Watch out: multiply m and Δθ before you divide when you rearrange for c. The bracket matters: E / m / Δθ is not the same as E / (m x Δθ) unless you do the steps in order.
Exam tip
Exam tip. Write the rearranged equation first, then substitute. Keep the units on every line.
Exercise 3C Rearranging to find the rise, the mass and the specific heat capacity
Method. Use Δθ = E / (m x c).
- Divide. Δθ = 42000 / (1 x 4200) = 42000 / 4200 = 10 °C.42000 / 4200 = 10
- Write the answer. 10 °C
Check. 4200 x 10 = 42000. ✓
Method. Use Δθ = E / (m x c).
- Divide. Δθ = 63000 / (3 x 4200) = 63000 / 12600 = 5 °C.63000 / 12600 = 5
- Write the answer. 5 °C
Check. 12600 x 5 = 63000. ✓
Method. Use Δθ = E / (m x c).
- Divide. Δθ = 3850 / (2 x 385) = 3850 / 770 = 5 °C.3850 / 770 = 5
- Write the answer. 5 °C
Check. 770 x 5 = 3850. ✓
Method. Use Δθ = E / (m x c).
- Divide. Δθ = 18000 / (0.5 x 900) = 18000 / 450 = 40 °C.18000 / 450 = 40
- Write the answer. 40 °C
Check. 450 x 40 = 18000. ✓
Method. Use m = E / (c x Δθ).
- Divide. m = 84000 / (4200 x 10) = 84000 / 42000 = 2 kg.84000 / 42000 = 2
- Write the answer. 2 kg
Check. 2 x 4200 x 10 = 84000. ✓
Method. Use m = E / (c x Δθ).
- Divide. m = 9000 / (450 x 20) = 9000 / 9000 = 1 kg.9000 / 9000 = 1
- Write the answer. 1 kg
Check. 1 x 450 x 20 = 9000. ✓
Method. Use c = E / (m x Δθ).
- Divide. c = 4500 / (2 x 5) = 4500 / 10 = 450 J/kg °C.4500 / 10 = 450
- Write the answer. 450 J/kg °C
Check. 2 x 450 x 5 = 4500. ✓
Method. Use c = E / (m x Δθ).
- Divide. c = 13500 / (1.5 x 10) = 13500 / 15 = 900 J/kg °C.13500 / 15 = 900
- Write the answer. 900 J/kg °C
Check. 1.5 x 900 x 10 = 13500. ✓
Method. Rearrange to Δθ = E / (m c).
- Multiply. m x c = 0.6 x 4200 = 2520.2520
- Divide. Δθ = 30240 / 2520 = 12 °C.30240 / 2520 = 12
- Write the answer. 12 °C
Check. 0.6 x 4200 x 12 = 30240 J. ✓
Method. Find c, then match it to the table.
- Find c. c = 4500 / (1 x 5) = 900 J/kg °C.900
- Match. A value of 900 J/kg °C matches aluminium in the table.aluminium
- Write the answer. aluminium
Check. Iron has 450, which is half of 900. ✓
Method. Rearrange the equation.
- Rearrange. Δθ = E / (m x c).divide by c
- Link. A bigger c is in the bottom line, so Δθ is smaller.smaller
- Write the answer. Accept any answer that says E = m c Δθ means a bigger c gives a smaller Δθ.
Check. ✓
Method. Use m = E / (c x Δθ).
- Divide. m = 27000 / (900 x 30) = 27000 / 27000 = 1 kg.27000 / 27000 = 1
- Write the answer. 1 kg
Check. 1 x 900 x 30 = 27000. ✓
Method. Use c = E / (m x Δθ).
- Multiply. m x Δθ = 0.25 x 20 = 5.5
- Divide. c = 2250 / 5 = 450 J/kg °C.2250 / 5 = 450
- Write the answer. 450 J/kg °C
Check. 0.25 x 450 x 20 = 2250. ✓
Practical: specific heat capacity of a metal block
In the practical, a metal block of known mass has a hole with a heater inside it. The block is wrapped in insulation to reduce heat loss. A thermometer is placed in a second hole. The heater is switched on for a known time, and the energy from the heater is found from its voltage and current: E = V x I x t. The temperature is recorded before and after heating. The specific heat capacity is then c = E / (m x Δθ).
Safety: the heater is low-voltage, but the practical is done only with a teacher or a grown-up. Do not touch the block while it is hot.
| Quantity | Value |
|---|---|
| start temperature | 20 degC |
| end temperature | 60 degC |
| temperature rise | 40 degC |
| energy from heater (V x I x t) | 9000.0 J |
| mass of block | 0.5 kg |
| specific heat capacity (E / m x rise) | 450 J/kg degC |
Key ideas
- Energy from the heater: E = V x I x t, in joules.
- Specific heat capacity: c = E / (m x Δθ).
- Use the mass of the block in kilograms.
- Insulation and a lid reduce the heat lost, so the result is more accurate.
Method. Use E = V x I x t.
- Power. Power = V x I = 12 x 2.5 = 30 W.30 W
- Energy. E = P x t = 30 x 300 = 9000 J.9000 J
- Write the answer. 9000 J
Check. 9000 J / 300 s = 30 W. ✓
Method. Use c = E / (m x Δθ).
- Multiply. m x Δθ = 0.5 x 40 = 20.20
- Divide. c = 9000 / 20 = 450 J/kg °C.9000 / 20 = 450
- Write the answer. 450 J/kg °C
Check. The value matches iron in the table. ✓
Method. Think about where the energy goes.
- Think. If some of the heater’s energy escapes to the surroundings, the block gains less energy than the heater supplied.energy lost
- Link. The calculation uses the heater energy, so it overestimates the energy absorbed and gives a value that is too high.overestimate
- Write the answer. Accept any sentence that says heat is lost to the surroundings, so the calculated energy is too large.
Check. More insulation would reduce this error. ✓
Watch out
Watch out: the energy used in the calculation is the energy from the heater. Heat lost to the surroundings means the block gains less than this, so c is overestimated unless the loss is reduced.
Exam tip
Practical safety. Always use the low-voltage supply, keep wires away from water, and let the block cool before handling it.
Exercise 3D The practical and the effect of heat loss
Method. Think about what the insulation does.
- Recall. Insulation reduces heat lost to the surroundings, so the block gains more of the heater energy.reduces heat loss
- Write the answer. reduce
Check. Insulation reduces heat loss. ✓
Method. Use E = V x I x t.
- Multiply. E = 12 x 2.5 x 300 = 9000 J.12 x 2.5 x 300 = 9000
- Write the answer. 9000 J
Check. 9000 / 300 = 30 W, and 12 x 2.5 = 30. ✓
Method. Look at the apparatus.
- Recall. The thermometer is placed in a hole in the block to read the temperature.thermometer
- Write the answer. thermometer
Check. ✓
Method. Think about the equation for c.
- Recall. c = E / (m x Δθ), so the mass is needed.m is needed
- Write the answer. Accept any sentence that says the mass is needed in the calculation of c.
Check. ✓
Method. Subtract the start temperature from the end temperature.
- Subtract. Δθ = 60 - 20 = 40 °C.60 - 20 = 40
- Write the answer. 40 °C
Check. 20 + 40 = 60. ✓
Method. Use E = m c Δθ with c = 450 J/kg °C.
- Multiply. E = 0.4 x 450 x 20 = 3600 J.0.4 x 450 x 20 = 3600
- Write the answer. 3600 J
Check. 3600 / (0.4 x 20) = 450. ✓
Method. Use E = P x t.
- Multiply. E = 40 x 150 = 6000 J.40 x 150 = 6000
- Write the answer. 6000 J
Check. 6000 / 150 = 40 W. ✓
Method. Use c = E / (m x Δθ).
- Multiply. m x Δθ = 0.5 x 24 = 12.12
- Divide. c = 6000 / 12 = 500 J/kg °C.6000 / 12 = 500
- Write the answer. 500 J/kg °C
Check. 0.5 x 500 x 24 = 6000 J. ✓
Method. Think about the energy used in the calculation.
- Supplied. The calculation uses the energy supplied by the heater.supplied
- Absorbed. Less energy is absorbed because of the heat loss, so the calculated c is too large.too large
- Write the answer. Accept any answer that says heat escapes to the surroundings, so the energy absorbed by the block is smaller than the energy supplied.
Check. ✓
Method. Think about the sources of error.
- Heat loss. Use insulation and a lid to reduce heat loss.insulation, lid
- Reading. Repeat the readings and find a mean, using a thermometer with fine divisions.repeat, mean
- Write the answer. Accept any two of: use insulation, use a lid, use a thermometer with finer divisions, repeat and find the average, or wait until the temperature is steady.
Check. ✓
Method. Think about what the error does to the temperature rise.
- Rise. The rise is too small by 2 °C.rise too small
- Decide. c = E / (m x rise), so a smaller rise gives a bigger c, which is too high.c too high
- Write the answer. high
Check. ✓
Why water is used in heating and cooling
Because water has a high specific heat capacity, a given mass of water can store a lot of thermal energy for each degree of rise. It is cheap, safe and easy to pump round a system. Cooling systems in cars also use water for the same reason: it can take in a lot of energy with only a small rise in temperature.
Key ideas
- Water: high c, so it stores a lot of energy for a given rise.
- A high c means the temperature changes slowly when energy is added or removed.
- Water is used in heating and cooling systems for this reason.
- Coastal areas have milder temperatures because the sea warms and cools slowly.
Method. Use E = m c Δθ with c = 4200 J/kg °C for water.
- Take the numbers. m = 1 kg, c = 4200 J/kg °C, Δθ = 10 °C.m = 1, c = 4200, Δθ = 10
- Multiply. E = 1 x 4200 x 10 = 42000 J.1 x 4200 x 10 = 42000
- Write the answer. 42000 J
Check. For the iron, E = 1 x 450 x 10 = 4500 J. ✓
Method. Divide the two energies.
- Iron. E for iron = 1 x 450 x 10 = 4500 J.4500 J
- Divide. Water / iron = 42000 / 4500.42000 / 4500
- Simplify. Divide both by 4500: 42000 / 4500 = 9.33 (to 3 significant figures), which is 9.33 times.9.33
- Write the answer. 9.33
Check. 4200 / 450 = 9.33 to 3 significant figures. ✓
Method. Use the idea of specific heat capacity.
- Think. The water must take in a lot of thermal energy from the engine.takes in energy
- Link. Because c is high, the water stays cooler while it absorbs the energy.stays cooler
- Write the answer. Accept any sentence that says water has a high specific heat capacity, so it absorbs a lot of energy with a small rise in temperature.
Check. ✓
Watch out
Watch out: a high specific heat capacity does not mean “hot”. It means the material changes temperature slowly for a given energy.
Exam tip
Exam tip. In an explanation question, link the property to the effect: “high c, so it stores more energy for the same rise”.
Exercise 3E Water, heating and cooling
Method. Think about the specific heat capacity of water.
- Recall. Water has a high specific heat capacity, so it stores a lot of energy.high c
- Write the answer. high
Check. ✓
Method. Use E = m c Δθ.
- Multiply. E = 1 x 4200 x 25 = 105000 J.1 x 4200 x 25 = 105000
- Write the answer. 105000 J
Check. 105000 / 25 = 4200. ✓
Method. Use E = m c Δθ with c = 385 J/kg °C.
- Multiply. E = 10 x 385 x 10 = 38500 J.10 x 385 x 10 = 38500
- Write the answer. 38500 J
Check. 38500 / 100 = 385. ✓
Method. The lower specific heat capacity warms more.
- Compare. Copper has the lower c, so it warms more.385 < 4200
- Write the answer. copper
Check. Copper warms about 11 times more. ✓
Method. Rearrange E = m c Δθ.
- Divide. Δθ = 84000 / (2 x 4200) = 10 °C.84000 / 8400 = 10
- Write the answer. 10 °C
Check. ✓
Method. Think about the energy stored in the water.
- Store. The water stores a lot of energy because c is high.stores
- Cool. It gives out this energy slowly, so the bottle cools slowly.cools slowly
- Write the answer. Accept any sentence that says water has a high specific heat capacity, so it cools slowly.
Check. ✓
Method. More energy for the same mass gives a bigger temperature rise.
- Compare. Mass 1 has the bigger rise, as Δθ = E / (m c).bigger rise
- Write the answer. 1
Check. ✓
Method. Compare the specific heat capacities.
- Compare. The metal has a lower c than water, so it heats up faster, not slower.metal lower c
- Write the answer. no
Check. Water takes longer to heat up. ✓
Method. Use E = m c Δθ with Δθ = 20 °C.
- Rise. The fall is 80 - 60 = 20 °C.20
- Multiply. E = 2 x 450 x 20 = 18000 J.2 x 450 x 20 = 18000
- Write the answer. 18000 J
Check. 18000 / (2 x 20) = 450. ✓
Method. Compare the sea and the land.
- Sea. Water has a high specific heat capacity, so the sea needs a lot of energy to warm.high c
- Land. Land has a lower specific heat capacity, so it warms faster in summer and cools faster in winter.lower c
- Write the answer. Accept any answer that says water has a high specific heat capacity so it takes longer to warm up and cool down.
Check. ✓
Chapter summary
| Skill | Method |
|---|---|
| Define specific heat capacity | Energy needed to warm 1 kg by 1 °C; unit J/kg °C |
| Calculate energy | E = m x c x Δθ, with m in kg |
| Rearrange | Divide E by the other two quantities multiplied together |
| Find c in the practical | E = V x I x t, then c = E / (m x Δθ) |
| Evaluate heat loss | Heat lost to the surroundings makes c too high unless insulation is used |
Can you…?
- I can define specific heat capacity and give its unit
- I can calculate the energy needed to warm a material
- I can rearrange E = mcΔθ to find the mass, rise or c
- I can describe the practical and the energy from a heater
- I can explain how heat loss affects the value of c
Key terms
- specific heat capacity (c)
- the energy needed to raise the temperature of 1 kg of a material by 1 °C
- temperature change (Δθ)
- the final temperature minus the starting temperature, in °C
- insulation
- a material that slows the transfer of thermal energy
- low-voltage supply
- a power supply that gives a safe, low voltage, used with a teacher
Exercise 3Z Check-up
Method. Energy per kilogram per degree.
- Recall. J/kg °C.J/kg °C
- Write the answer. J/kg °C
Check. ✓
Method. Use E = m c Δθ.
- Multiply. E = 4 x 4200 x 5 = 84000 J.4 x 4200 x 5 = 84000
- Write the answer. 84000 J
Check. ✓
Method. Use E = m c Δθ.
- Multiply. E = 0.5 x 450 x 20 = 4500 J.0.5 x 450 x 20 = 4500
- Write the answer. 4500 J
Check. ✓
Method. Rearrange to Δθ = E / (m c).
- Divide. 42000 / 4200 = 10 °C.42000 / 4200 = 10
- Write the answer. 10 °C
Check. ✓
Method. Use m = E / (c Δθ).
- Divide. m = 18000 / (900 x 20) = 18000 / 18000 = 1 kg.18000 / 18000 = 1
- Write the answer. 1 kg
Check. ✓
Method. Use c = E / (m Δθ).
- Divide. c = 9000 / (2 x 5) = 900 J/kg °C.9000 / 10 = 900
- Write the answer. 900 J/kg °C
Check. ✓
Method. Use E = V x I x t.
- Multiply. E = 12 x 2 x 100 = 2400 J.12 x 2 x 100 = 2400
- Write the answer. 2400 J
Check. ✓
Method. Think about where the energy goes.
- Reason. Heat is lost to the surroundings, so the block gains less than the heater supplies.heat lost
- Write the answer. heat is lost to the surroundings
Check. ✓
Method. Check the value for water.
- Check. Water has a high specific heat capacity, so the pupil is wrong.high c
- Write the answer. no
Check. ✓
Method. Use Δθ = E / (m c).
- Divide. 3850 / (3 x 385) = 3850 / 1155 = 3.33 °C.3850 / 1155
- Write the answer. 3.33 °C
Check. ✓
Total 31 marks. Allow about 31 minutes.
Exercise 3X Exam-style questions
Method. Use E = m c Δθ.
- Multiply. E = 2 x 385 x 10 = 7700 J.2 x 385 x 10 = 7700
- Write the answer. 7700 J
Check. ✓
Method. Use Δθ = E / (m c).
- Divide. Δθ = 3600 / (0.6 x 450) = 3600 / 270 = 13.3 °C to 1 decimal place.3600 / 270 = 13.3
- Write the answer. 13.3 °C
Check. ✓
Method. Use c = E / (m Δθ).
- Multiply. m x Δθ = 0.2 x 30 = 6.6
- Divide. c = 1800 / 6 = 300 J/kg °C.1800 / 6 = 300
- Write the answer. 300 J/kg °C
Check. ✓
Method. Find the energy first, then c.
- Energy. E = 12 x 3 x 200 = 7200 J.7200 J
- Divide. c = 7200 / (1 x 8) = 900 J/kg °C.7200 / 8 = 900
- Write the answer. 900 J/kg °C
Check. ✓
Method. Heater energy first, then the energy the water gains.
- Heater. E = 60 x (5 x 60) = 60 x 300 = 18000 J.18000 J
- Water. E = 0.5 x 4200 x 8 = 16800 J.16800 J
- Write the answer. 18000, 16800
Check. The water gains less than the heater supplies, because some heat is lost. ✓
Method. Think about the energy in the calculation and the effect of heat loss.
- Loss. Heat is lost to the surroundings, so the block absorbs less than the heater supplies.heat lost
- Effect. The calculation uses the heater energy, so c comes out too high.too high
- Change. Use insulation around the block and a lid on the hole to reduce heat loss.insulation, lid
- Write the answer. Accept any answer that says heat is lost to the surroundings, and names insulation or a lid.
Check. ✓
Method. Find c, then match it to the table.
- Divide. c = 9000 / (1 x 25) = 360 J/kg °C.9000 / 25 = 360
- Match. The closest value is copper, 385 J/kg °C, which is nearest to 360.copper is nearest
- Write the answer. copper
Check. Accept the closest value. ✓
Method. Describe the method in order.
- Mass. Measure the mass of the block in kilograms.mass in kg
- Heat. Insert the heater and thermometer, wrap in insulation, and record the start temperature.start temperature
- Energy. Switch on the low-voltage supply for a known time, and record the voltage and current; E = V x I x t.E = V x I x t
- Rise. Record the final temperature and find the rise.temperature rise
- Calculate. Use c = E / (m x Δθ).c = E / (m x change in theta)
- Write the answer. Accept any answer that uses E = V x I x t, measures mass and temperature rise, and uses c = E / (m x Δθ).
Check. ✓
Method. Weigh the properties.
- Property. Water has a high specific heat capacity, so it stores a lot of energy for each degree.high c
- Cost. Cost is a secondary point; it does not explain why water stores energy well.secondary
- Conclusion. So the claim is partly right, but the high specific heat capacity is the main reason.partly right
- Write the answer. Accept any answer that says the high specific heat capacity is the key property, and that cost is a secondary point.
Check. ✓
Method. Use c = E / (m Δθ), then consider heat loss.
- Multiply. m x Δθ = 4 x 15 = 60.60
- Divide. c = 16200 / 60 = 270 J/kg °C.16200 / 60 = 270
- Heat loss. If heat is lost, the energy absorbed is less than 16200 J, so the true c is less than 270.absorbed < supplied
- Write the answer. 270 J/kg °C
Check. 60 x 270 = 16200. ✓