Specific heat capacity

1 Plan and risk assessment

The specific heat capacity of a material is the energy needed to raise the temperature of 1 kg of it by 1 °C. Heat a block electrically, measure the energy with a joulemeter and the temperature rise with a thermometer, and c = ΔE ÷ (m Δθ).

ΔE = m c Δθ

Prediction

2 Apparatus

• ✓1 kg metal block with two holes The material being tested.
• ✓Top-pan balance (resolution 0.1 g) Measures the mass of the block.
• ✓Immersion heater (12 V) Transfers energy into the block.
• ✓Thermometer (−10 to 110 °C) (resolution 1 °C) Measures the temperature of the block. Its scale is marked every 1 °C; read it to the nearest 0.5 °C, halfway between two marks.
• ✓Joulemeter (resolution 1 J) Measures the energy transferred by the heater.
• ✓Low-voltage power supply Powers the heater.
• ✓Pipette and water A few drops in the thermometer hole for good thermal contact.
• ✓Insulation jacket Reduces energy transfer to the surroundings.
• ✓Stopwatch Times the readings.
• ✓Heatproof mat Protects the bench from the hot block and heater.

3 Method: your choices

• Which version are you doing?: AQA required practical 1: specific heat capacity of a metal blockHeat a 1 kg block with an immersion heater and a joulemeter for 10 minutes.
• Choose your block: AluminiumThe usual school block, and the largest of the three.
• Add water to the thermometer hole: Put a few drops of water in the hole with the pipetteWater fills the gap round the bulb, so the thermometer keeps up with the block and the graph straightens within a minute or two.
• Wrap the block in insulation: Wrap the insulation jacket round the blockLess energy escapes to the room, so more of it warms the block. Your value of c is still a little too high, but closer.

• 0.0 Ringed: the anomaly, left out of your results.

Anomaly: Temperature at 360 s, left out of your results. Likely cause: the thermometer was lifted out of the hole, so for a moment it was not measuring the block.

6 Analysis

1. Mass of the block

   1 kg = 1000 g

   The balance read 1001.5 g when you weighed the block. Write its mass in kilograms.

   1.0015 kg Right first time

2. Shape of the graph

   What shape is your graph?

   A short curve at the start, then a straight line. Right first time

3. Gradient (temperature rise per joule)

   gradient = change in y ÷ change in x

   0.00102 °C/J Right first time

4. Specific heat capacity

   ΔE = m c Δθ, so c = 1 ÷ (m × gradient)

   Your gradient is 0.00102 °C/J and the mass of the block is 1.0015 kg. Calculate the specific heat capacity of the metal.

   980 J/kg °C Right first time

5. Compare with the data book

   The data-book value for aluminium is 900 J/kg °C. How does your value compare?

   Higher than the data-book value Right first time

7 Conclusion and evaluation

Conclusion

Sources of error

Improvement

Repeatable or reproducible

9 Exam questions on your data

On your own readings, self-marked against the mark scheme, with error carried forward where it applies. Total 30/30.

1. 01.1[2 marks]

   You did an experiment to determine the specific heat capacity of aluminium.

   An immersion heater was placed in one hole in a 1.002 kg aluminium block and a thermometer in the other. A joulemeter measured the energy transferred by the heater. You recorded the temperature of the block and the joulemeter reading every 60 s for 600 s.

   Which two statements about the variables in this investigation are correct?

   Tick (✓) two boxes.

   • The energy transferred to the block is a control variable.
   • ✓ The independent variable is the energy transferred to the block. (your choice)
   • The dependent variable is the mass of the block.
   • The independent variable is the temperature of the block.
   • ✓ The dependent variable is the temperature of the block. (your choice)

   Your answer

   The independent variable is the energy transferred to the block.; The dependent variable is the temperature of the block.

   2/2

   • The independent variable is the energy transferred to the block.
   • The dependent variable is the temperature of the block.

   Full marks. That is exactly what the mark scheme is looking for.

   Model answer: The independent variable is the energy transferred to the block. The dependent variable is the temperature of the block.

2. 01.2[1 mark]

   Which statement gives the definition and the unit of specific heat capacity?

   Tick (✓) one box.

   • The energy needed to raise the temperature of the substance by 1 °C. Unit: J/°C
   • ✓ The energy needed to raise the temperature of 1 kg of the substance by 1 °C. Unit: J/kg °C (your choice)
   • The energy needed to change the state of 1 kg of the substance with no change in temperature. Unit: J/kg
   • The energy needed to raise the temperature of 1 kg of the substance by 1 °C. Unit: J/kg

   Your answer

   The energy needed to raise the temperature of 1 kg of the substance by 1 °C. Unit: J/kg °C

   1/1

   • The energy needed to raise the temperature of 1 kg of the substance by 1 °C. Unit: J/kg °C

   Full marks. Your answer meets all the points for this question.

   Model answer: The energy needed to raise the temperature of 1 kg of the substance by 1 °C. Unit: J/kg °C

3. 01.3[2 marks]

   You wrapped the block in insulation before switching on the heater.

   Explain why.

   Your answer

   Insulation reduces the energy transferred from the block to the surroundings, so more of the energy from the heater increases the temperature of the block. This makes the value of the specific heat capacity more accurate.

   2/2 self-marked

   • To reduce energy transferred (lost) to the surroundings
   • So more of the energy from the heater increases the temperature of the block / the value of c is more accurate (ignore "more precise")

   Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

   Model answer: Insulation reduces the energy transferred from the block to the surroundings, so more of the energy from the heater increases the temperature of the block. This makes the value of the specific heat capacity more accurate.

4. 01.4[2 marks]

   You put a few drops of water into the hole before putting the thermometer in.

   Explain why.

   Your answer

   The water improves the thermal contact between the block and the thermometer, because water is a better thermal conductor than air. So the thermometer reading is the same as the temperature of the block.

   2/2 self-marked

   • Improves thermal contact between the thermometer and the block (water is a better conductor than air)
   • So the thermometer reads the temperature of the block (more accurate reading, responds faster)

   Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

   Model answer: The water improves the thermal contact between the block and the thermometer, because water is a better thermal conductor than air. So the thermometer reading is the same as the temperature of the block.

5. 01.5[1 mark]

   A student put the thermometer into the aluminium block. The student waited a few minutes before reading the starting temperature.

   Suggest why the student waited before reading the starting temperature.

   Your answer

   So that the thermometer reaches the same temperature as the block, and the starting reading is the temperature of the block.

   1/1 self-marked

   • So the thermometer reaches the same temperature as the block (it must refer to the block, not the room)

   Self-marked: 1 of 1 mark. This is your own judgement against the mark scheme.

   Model answer: So that the thermometer reaches the same temperature as the block, and the starting reading is the temperature of the block.

6. 01.6[1 mark]

   Another student used a digital thermometer. At the start of the experiment it showed 22.2 °C.

   Give the resolution of the digital thermometer.

   Your answer

   0.1

   1/1

   • 0.1 °C

   Full marks. Well done.

   Model answer: 0.1 °C

7. 01.7[1 mark]

   A student heats a metal block with an electric immersion heater. A thermometer is placed in a second hole in the block.

   Give one hazard in this investigation.

   Your answer

   The heater and the block get hot, so you could get burns if you touch them.

   1/1 self-marked

   • Burns (from the hot heater or block); allow cuts from a broken glass thermometer. Ignore "the heater is hot" on its own.

   Self-marked: 1 of 1 mark. This is your own judgement against the mark scheme.

   Model answer: The heater and the block get hot, so you could get burns if you touch them.

8. 02.1[2 marks]

   Table 1 shows your results.

   In the first 60 s the temperature of the block increased by only 0.5 °C. Between 120 s and 180 s it increased by 1.5 °C.

   Table 1

   Time in s	Energy transferred in J	Temperature in °C
   0	0	22.0
   60	1643	22.5
   120	3286	24.0
   180	4929	25.5
   240	6572	27.5
   300	8216	29.0
   360	9859	29.0
   420	11 502	32.5
   480	13 145	34.0
   540	14 788	35.5
   600	16 432	37.5

   Your teacher said you should not use the first two readings when drawing the line of best fit to find the gradient.

   Explain why.

   Your answer

   The first readings do not follow the steady rise of the later readings (the temperature rises more slowly at the start), so they are not on the straight part of the graph. This is because the heater has to warm up itself and the energy takes time to be conducted through the block to the thermometer.

   2/2 self-marked

   • The first readings do not follow the steady rise (the temperature rises more slowly at the start), so they are not on the straight part of the graph (allow the temperature does not increase steadily at the start; the first readings do not fit the pattern)
   • The heater had to warm up itself first / energy takes time to be conducted through the block to the thermometer

   Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

   Model answer: The first readings do not follow the steady rise of the later readings (the temperature rises more slowly at the start), so they are not on the straight part of the graph. This is because the heater has to warm up itself and the energy takes time to be conducted through the block to the thermometer.

9. 02.2[1 mark]

   Table 1 shows your results again.

   One of the temperature readings in Table 1 is anomalous.

   Identify the anomalous reading.

   Select one row in Table 1.

   Table 1

   Time in s	Energy transferred in J	Temperature in °C
   0	0	22.0
   60	1643	22.5
   120	3286	24.0
   180	4929	25.5
   240	6572	27.5
   300	8216	29.0
   360	9859	29.0
   420	11 502	32.5
   480	13 145	34.0
   540	14 788	35.5
   600	16 432	37.5

   Your answer

   the reading at 360 s (29.0 °C)

   1/1

   • The reading at 360 s (29.0 °C)

   Full marks. That is exactly what the mark scheme is looking for. It is lower than the steady pattern of the readings either side, so it should be ignored when drawing the line of best fit.

   Model answer: The reading at 360 s (29.0 °C). It does not fit the steady increase shown by the readings either side of it.

10. 02.3[4 marks]

    The mass of the block was 1.002 kg.

    Table 1

    Time in s	Energy transferred in J	Temperature in °C
    0	0	22.0
    60	1643	22.5
    120	3286	24.0
    180	4929	25.5
    240	6572	27.5
    300	8216	29.0
    360	9859	29.0
    420	11 502	32.5
    480	13 145	34.0
    540	14 788	35.5
    600	16 432	37.5

    change in thermal energy = mass × specific heat capacity × temperature change ΔE = m c Δθ

    Use your readings at 120 s and 600 s in Table 1 to calculate the specific heat capacity of aluminium.

    Use the Physics Equations Sheet.

    Your answer

    972

    Working: ΔE = 16 432 − 3286 = 13 146 J Δθ = 37.5 − 24.0 = 13.5 °C 13 146 = 1.002 × c × 13.5 c = 13 146 ÷ (1.002 × 13.5)

    4/4

    • ΔE = 16 432 − 3286 = 13 146 J
    • Δθ = 37.5 − 24.0 = 13.5 °C
    • Substitution and rearrangement: c = 13 146 ÷ (1.002 × 13.5)
    • c = 972 J/kg °C (allow error carried forward)

    Full marks. Your answer meets all the points for this question.

    Model answer: ΔE = 16 432 − 3286 = 13 146 J Δθ = 37.5 − 24.0 = 13.5 °C 13 146 = 1.002 × c × 13.5 c = 13 146 ÷ (1.002 × 13.5) c = 972 J/kg °C

11. 02.4[5 marks]

    Another student did the experiment with a 1.00 kg iron block. The student did not have a joulemeter.

    The power of the heater was 29 W. The heater was switched on at 0 s.

    Table 2 shows the student’s results.

    Table 2

    Time in s	Temperature in °C
    0	17.5
    60	19.5
    120	23.0
    180	26.5
    240	30.0
    300	33.5
    360	37.0
    420	40.5
    480	44.0
    540	47.5
    600	51.0

    change in thermal energy = mass × specific heat capacity × temperature change ΔE = m c Δθ

    Use the readings at 120 s and 600 s in Table 2 to calculate the specific heat capacity of iron.

    Use the Physics Equations Sheet.

    Give your answer to 2 significant figures.

    Your answer

    500

    Working: E = 29 × 480 E = 13 920 J Δθ = 51.0 − 23.0 = 28.0 °C 13 920 = 1.00 × c × 28.0 c = 13 920 ÷ (1.00 × 28.0) c = 497.1

    5/5

    • E = 29 × 480
    • E = 13 920 J
    • 13 920 = 1.00 × c × 28.0 or c = 13 920 ÷ (1.00 × 28.0) (allow their E and their Δθ; E = P t must have been used)
    • c = 497.1 (allow an answer consistent with their E)
    • c = 500 J/kg °C (2 significant figures)

    Full marks. That is exactly what the mark scheme is looking for.

    Model answer: E = 29 × 480 E = 13 920 J Δθ = 51.0 − 23.0 = 28.0 °C 13 920 = 1.00 × c × 28.0 c = 13 920 ÷ (1.00 × 28.0) c = 497.1 c = 500 J/kg °C (2 significant figures)

12. 02.5[2 marks]

    Your value from your graph for the specific heat capacity of aluminium was 980 J/kg °C.

    The accepted value is 900 J/kg °C.

    Explain why your value is higher than the accepted value.

    Your answer

    Even with insulation, some energy from the heater is transferred to the surroundings (and some heats the heater itself). So the temperature rise of the block is smaller than it would be if all the energy heated the block, and c = ΔE ÷ (m × Δθ) comes out too big.

    2/2 self-marked

    • Energy is transferred (lost) to the surroundings, or heats the heater / thermometer
    • So the temperature rise of the block is smaller than expected (so ΔE ÷ mΔθ is bigger)

    Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

    Model answer: Even with insulation, some energy from the heater is transferred to the surroundings (and some heats the heater itself). So the temperature rise of the block is smaller than it would be if all the energy heated the block, and c = ΔE ÷ (m × Δθ) comes out too big.

13. 03[6 marks]

    Describe a method a student could use to determine the specific heat capacity of a metal block.

    You should include how the student would use the results to find the specific heat capacity.

    Your answer

    1. Measure the mass of the aluminium block using a top-pan balance. 2. Put the immersion heater into the large hole and the thermometer into the small hole. Add a few drops of water to the thermometer hole to give good thermal contact. 3. Wrap the block in insulation to reduce energy transfer to the surroundings. 4. Connect the heater to the power supply through a joulemeter. 5. Record the starting temperature of the block. 6. Switch on the power supply and start the stopwatch. 7. Record the temperature (reading the thermometer at eye level) and the joulemeter reading every minute for 10 minutes. 8. Plot a graph of temperature (y-axis) against energy transferred (x-axis) and draw a line of best fit through the straight part. 9. Find the gradient of the line. Specific heat capacity = 1 ÷ (mass × gradient). 10. The heater gets very hot, so do not touch it until it has cooled down.

    6/6 self-marked

    • Measure the mass of the block with a top-pan balance
    • Put the immersion heater in the hole in the block
    • Put the thermometer in the other hole
    • Add a few drops of water to the thermometer hole for good thermal contact
    • Wrap the block in insulation
    • Connect the heater to a power supply and joulemeter
    • Measure the temperature of the block with a thermometer
    • Measure the starting temperature of the block
    • Switch on the heater and start the stopwatch
    • Record the temperature at regular intervals, e.g. every minute
    • Record the energy transferred (joulemeter reading) each time
    • Take readings for about 10 minutes
    • Plot a graph of temperature against energy transferred
    • Find the gradient of the straight part of the line of best fit
    • Specific heat capacity = 1 ÷ (mass × gradient), or use ΔE = m c Δθ
    • Safety: the heater gets hot, so do not touch it until it has cooled
    • Repeat the experiment and calculate a mean

    Self-marked: Level 3, 6 of 6 marks. You ticked 17 of 17 indicative points. This is your own judgement against the mark scheme.

    Model answer: 1. Measure the mass of the aluminium block using a top-pan balance. 2. Put the immersion heater into the large hole and the thermometer into the small hole. Add a few drops of water to the thermometer hole to give good thermal contact. 3. Wrap the block in insulation to reduce energy transfer to the surroundings. 4. Connect the heater to the power supply through a joulemeter. 5. Record the starting temperature of the block. 6. Switch on the power supply and start the stopwatch. 7. Record the temperature (reading the thermometer at eye level) and the joulemeter reading every minute for 10 minutes. 8. Plot a graph of temperature (y-axis) against energy transferred (x-axis) and draw a line of best fit through the straight part. 9. Find the gradient of the line. Specific heat capacity = 1 ÷ (mass × gradient). 10. The heater gets very hot, so do not touch it until it has cooled down.

Unseen data: the same skills, new apparatus

1. 01[3 marks]

   A student heats a 0.8 kg block of a new alloy with a 40 W heater and reads its temperature every minute. The block is wrapped in insulation.

   Temperature of the block

   Time in min	Temperature in °C
   0	18.9
   1	20.6
   2	24.1
   3	27.9
   4	32.0
   5	35.9
   6	39.9

   Calculate the mean rate of temperature rise of the block between 2 and 6 minutes, in °C per minute. Give your answer to 2 significant figures.

   Your answer

   4.0

   Working: 39.9 - 24.1 = 15.8 6 - 2 = 4 15.8 / 4

   3/3

   • Rise = 39.9 − 24.1 = 15.8 °C.
   • Time taken = 6 − 2 = 4 min.
   • Mean rate = rise ÷ time taken, in °C per minute.

   Full marks. Clear and precise.

   Model answer: Rise = 39.9 − 24.1 = 15.8 °C in 6 − 2 = 4 minutes. Mean rate of rise = 15.8 ÷ 4 = 4.0 °C per minute.

2. 02[2 marks]

   A student measures the specific heat capacity of a zinc block twice with the same heater and joulemeter: once with the block bare, once wrapped in a thick insulating jacket.

   Specific heat capacity of the block

   	Value in J/kg °C
   Bare block	475
   Wrapped block	410
   Data book	390

   Which TWO statements about wrapping the block are correct? Tick (✓) two boxes.

   • ✓ The value with the jacket is more accurate: it is closer to the data-book value. (your choice)
   • ✓ The jacket lowers the rate of energy transfer from the block to the surroundings. (your choice)
   • The jacket lowers the power of the heater.
   • With the jacket the block’s temperature rises more slowly.
   • The jacket makes the uncertainty in each temperature reading bigger.

   Your answer

   The value with the jacket is more accurate: it is closer to the data-book value.; The jacket lowers the rate of energy transfer from the block to the surroundings.

   2/2

   • The value with the jacket is more accurate: it is closer to the data-book value.
   • The jacket lowers the rate of energy transfer from the block to the surroundings.

   Full marks. Clear and precise.

   Model answer: The jacket lowers the rate of energy transfer to the surroundings, so more of the energy warms the block: its temperature rises more (and faster) and c comes out at 410 J/kg °C, closer to the data book’s 390 than the bare block’s 475. The heater’s power and the thermometer are unchanged.

3. 03[2 marks]

   An engineer chooses a liquid to carry energy away from a hot motor. The same mass of liquid flows past the motor each second.

   Specific heat capacity of each liquid

   Liquid	Specific heat capacity in J/kg °C
   A	3300
   B	2100
   C	4200

   Which TWO statements are correct? Tick (✓) two boxes.

   • ✓ Liquid C is the best choice. (your choice)
   • ✓ A liquid with a high specific heat capacity can take in a lot of energy with only a small rise in temperature. (your choice)
   • Liquid B is the best choice, because it warms up fastest.
   • All three need the same energy to warm 1 kg of liquid by 1 °C.

   Your answer

   Liquid C is the best choice.; A liquid with a high specific heat capacity can take in a lot of energy with only a small rise in temperature.

   2/2

   • Liquid C is the best choice.
   • A liquid with a high specific heat capacity can take in a lot of energy with only a small rise in temperature.

   Full marks. Your answer meets all the points for this question.

   Model answer: Liquid C has the highest specific heat capacity (4200 J/kg °C): each kilogram takes in the most energy for each degree it warms, so it carries the most energy away while staying coolest.