P4: Specific Heat Capacity
Learn about specific heat capacity, how to calculate the energy needed to change temperature, and how to carry out the required practical investigation.
What is Specific Heat Capacity?
Different materials need different amounts of energy to change their temperature. Some materials heat up quickly, while others take a long time. This property is called specific heat capacity.
Definition
Specific heat capacity is the amount of energy required to raise the temperature of 1 kg of a substance by 1 degrees Celsius. It is measured in joules per kilogram per degree Celsius (J/kg degrees C).
Understanding SHC
A substance with a high specific heat capacity takes a lot of energy to change its temperature. This means it heats up slowly and cools down slowly. Water has a very high SHC (4200 J/kg degrees C), which is why oceans heat up and cool down slowly. A substance with a low specific heat capacity takes less energy to change its temperature. Metals like aluminium and copper have low SHC values, which is why they heat up quickly.
The Specific Heat Capacity Equation
Specific Heat Capacity Equation
E = mcΔT
E = energy transferred (J)
m = mass (kg)
c = specific heat capacity (J/kg degrees C)
ΔT = change in temperature (degrees C)
Rearranging the Equation
To find the SHC: c = E / (m × ΔT)
To find the mass: m = E / (c × ΔT)
To find the temperature change: ΔT = E / (m × c)
Make sure the mass is in kilograms, not grams. If the question gives mass in grams, divide by 1000 to convert to kilograms before using the equation.
Worked Examples
Worked Example 1
Calculate the energy required to heat 2 kg of water from 20 degrees C to 80 degrees C. The specific heat capacity of water is 4200 J/kg degrees C.
E = mcΔT
ΔT = 80 - 20 = 60 degrees C
E = 2 × 4200 × 60
E = 504 000 J
E = 504 kJ
Worked Example 2
500 g of aluminium is heated from 25 degrees C to 75 degrees C. The specific heat capacity of aluminium is 900 J/kg degrees C. Calculate the energy transferred.
First convert mass: m = 500 g = 0.5 kg
ΔT = 75 - 25 = 50 degrees C
E = mcΔT
E = 0.5 × 900 × 50
E = 22 500 J
E = 22.5 kJ
Worked Example 3
A heater provides 15 000 J of energy to 0.3 kg of copper. The specific heat capacity of copper is 385 J/kg degrees C. Calculate the temperature increase.
ΔT = E / (m × c)
ΔT = 15 000 / (0.3 × 385)
ΔT = 15 000 / 115.5
ΔT = 129.9 degrees C
ΔT = 130 degrees C (2 s.f.)
Worked Example 4
10 000 J of energy is supplied to 0.2 kg of an unknown metal. The temperature rises by 50 degrees C. Calculate the specific heat capacity of the metal.
c = E / (m × ΔT)
c = 10 000 / (0.2 × 50)
c = 10 000 / 10
c = 1000 J/kg degrees C
Worked Example 5
An electric kettle heats 1.5 kg of water from 18 degrees C to boiling (100 degrees C). The kettle has a power of 2000 W. Calculate the time taken for the water to boil. The SHC of water is 4200 J/kg degrees C.
Step 1: Calculate the energy needed
E = mcΔT = 1.5 × 4200 × (100 - 18)
E = 1.5 × 4200 × 82 = 516 600 J
Step 2: Calculate the time using power = energy / time
time = energy / power = 516 600 / 2000 = 258.3 s = 258 s (3 s.f.)
Worked Example 6
A 2 kg block of iron (SHC = 450 J/kg degrees C) at 200 degrees C is placed in 5 kg of water (SHC = 4200 J/kg degrees C) at 20 degrees C. Calculate the final temperature of the mixture, assuming no energy is lost to the surroundings.
Energy lost by iron = Energy gained by water
2 × 450 × (200 - T) = 5 × 4200 × (T - 20)
900(200 - T) = 21000(T - 20)
180000 - 900T = 21000T - 420000
180000 + 420000 = 21000T + 900T
600000 = 21900T
T = 27.4 degrees C
Comparison of Specific Heat Capacity Values
| Substance | Specific Heat Capacity (J/kg degrees C) | Implication |
|---|---|---|
| Water | 4200 | Very high - hard to heat up and cool down |
| Oil | 2000 | Moderate - heats faster than water |
| Ice | 2100 | About half that of liquid water |
| Aluminium | 900 | Low - heats up quickly |
| Iron | 450 | Low - heats up quickly |
| Copper | 385 | Low - heats up quickly |
| Lead | 130 | Very low - heats up very quickly |
| Granite | 790 | Low - used in storage heaters |
| Glass | 840 | Low - heats up fairly quickly |
| Air | 1005 | Moderate - important for climate |
Why Water is So Useful
Water has an extremely high specific heat capacity. This makes it very useful as a coolant in car engines and nuclear reactors because it can absorb a large amount of thermal energy without a large temperature rise. It is also why coastal areas have milder climates than inland areas: the sea absorbs heat in summer and releases it in winter, moderating the temperature.
Storage Heaters
Storage heaters use bricks or concrete blocks with a moderate specific heat capacity. They are heated overnight using cheaper off-peak electricity. During the day, they slowly release the stored thermal energy to warm the room. The moderate SHC means they store enough energy but release it gradually over many hours.
Required Practical: Measuring Specific Heat Capacity
Aim
To determine the specific heat capacity of a solid material (e.g. a metal block).
Equipment
Metal block of known mass, electric immersion heater, thermometer, power supply, ammeter, voltmeter, stopwatch, insulating material.
Method
1. Measure and record the mass of the metal block using a balance.
2. Insert the immersion heater and thermometer into the holes in the metal block.
3. Wrap the block in insulating material to reduce energy loss to the surroundings.
4. Record the initial temperature of the block.
5. Connect the heater to the power supply with an ammeter in series and a voltmeter in parallel.
6. Turn on the power supply and start the stopwatch simultaneously.
7. Record the ammeter reading (current, I) and voltmeter reading (voltage, V).
8. After a set time (e.g. 5 minutes), turn off the power supply and record the final temperature.
9. Calculate the energy supplied using: E = VIt (voltage × current × time).
10. Calculate the SHC using: c = E / (m × ΔT).
Sources of Error and Improvements
Not all the energy from the heater goes into the block. Some is lost to the surroundings. This means the calculated SHC will be higher than the true value because the actual energy going into the block is less than E = VIt. To improve accuracy: use thicker insulation around the block; ensure the heater and thermometer fit tightly in their holes; stir the block (if liquid) to distribute heat evenly; use a data logger for more accurate temperature readings.
In the required practical, the calculated value of SHC is usually HIGHER than the accepted value. This is because some energy is lost to the surroundings, so the actual energy gained by the block is less than the energy calculated from E = VIt. Since c = E/(mΔT), if E is too large (because we assumed all electrical energy went into the block), then c will be too high.
Practice Questions
1. Calculate the energy required to heat 3 kg of water from 15 degrees C to 85 degrees C. The SHC of water is 4200 J/kg degrees C.
E = mcΔT = 3 × 4200 × (85 - 15) = 3 × 4200 × 70 = 882 000 J = 882 kJ
2. A heater supplies 20 000 J of energy to 0.5 kg of aluminium. The SHC of aluminium is 900 J/kg degrees C. Calculate the temperature rise.
ΔT = E / (m × c) = 20 000 / (0.5 × 900) = 20 000 / 450 = 44.4 degrees C
3. 8000 J of energy heats 2 kg of a substance by 10 degrees C. Calculate its specific heat capacity.
c = E / (m × ΔT) = 8000 / (2 × 10) = 8000 / 20 = 400 J/kg degrees C
4. Explain why water is used as a coolant in car engines.
Water has a very high specific heat capacity (4200 J/kg degrees C). This means it can absorb a large amount of thermal energy from the engine without its temperature rising too much, making it an effective coolant.
5. In the SHC required practical, the calculated value of SHC was higher than the true value. Explain why and suggest how the experiment could be improved.
Some thermal energy is lost to the surroundings, so less energy actually heats the block than calculated from E = VIt. Since c = E/(mΔT), using a larger E gives a larger c. Improvements: use better insulation around the block, ensure heater and thermometer fit tightly, use a lid to reduce convection losses.
6. A 1500 W electric heater is used to heat 0.8 kg of oil (SHC = 2000 J/kg degrees C) from 22 degrees C to 65 degrees C. Calculate the minimum time the heater must be switched on, assuming no energy is lost.
E = mcΔT = 0.8 × 2000 × (65 - 22) = 0.8 × 2000 × 43 = 68 800 J. Time = E / power = 68 800 / 1500 = 45.9 s = 46 s (2 s.f.)
🔬 Required Practical
Required Practical: Investigating the Specific Heat Capacity of a Solid
Aim: To determine the specific heat capacity of a metal block by measuring the energy supplied electrically and the resulting temperature change.
Method: Measure the mass of the metal block using a balance. Insert an electric immersion heater and a thermometer into the holes in the block. Wrap the block in insulating material to reduce thermal energy loss to the surroundings. Record the initial temperature. Connect the heater to a DC power supply with an ammeter in series and a voltmeter in parallel across the heater. Switch on the power supply and start a stopwatch simultaneously. Record the ammeter and voltmeter readings. After a set time (e.g. 5 minutes), switch off the heater and record the final temperature. Calculate the energy supplied using E = VIt. Calculate the SHC using c = E / (mΔT).
Variables: Independent: time the heater is switched on (and therefore energy supplied, E = VIt). Dependent: temperature change of the block (ΔT). Control: mass of the block, type of material, insulation used, voltage and current settings.
Analysis: Plot a graph of temperature change against energy supplied. The gradient is equal to 1/(mc), so c = 1/(gradient × m). A straight line through the origin confirms the relationship E = mcΔT. The calculated SHC will typically be higher than the accepted value because some energy is lost to the surroundings — the actual energy going into the block is less than E = VIt.
Common exam questions: "Why is the calculated value of SHC higher than the true value?" — Because some thermal energy escapes to the surroundings, so less energy actually heats the block than is calculated from E = VIt. Since c = E/(mΔT), an overestimated E gives an overestimated c. "How could you improve the accuracy?" — Use thicker insulation, ensure the heater and thermometer fit tightly in their holes, use a lid, and use a data logger for more precise temperature readings.
🔢 Maths Skills
Mathematical Skills for this Topic
Rearranging E = mcΔT: To find the SHC: c = E / (m × ΔT). To find the mass: m = E / (c × ΔT). To find the temperature change: ΔT = E / (m × c). When rearranging, always write out the original equation first, then substitute known values before calculating. Avoid rounding intermediate values — carry extra significant figures through the calculation and round only the final answer.
Significant figures: Your answer should be given to the same number of significant figures as the least precise data value in the calculation. For example, if mass is given as 0.5 kg (1 s.f.) and temperature change is 50 °C (2 s.f.), give your answer to 1 s.f. In practice, most GCSE answers should be given to 2 or 3 significant figures. Always state the number of significant figures you have used.
Unit conversions (J to kJ): To convert joules to kilojoules, divide by 1000. To convert kilojoules to joules, multiply by 1000. Energy values involving SHC are often very large (e.g. 504 000 J), so converting to 504 kJ makes the answer clearer and less prone to errors. Also remember to convert grams to kilograms (divide by 1000) before using the SHC equation, and convert cm³ to m³ if necessary.
Calculating energy supplied electrically (E = VIt): In the required practical, you calculate energy using voltage, current, and time. Time must be in seconds — convert minutes by multiplying by 60. For example, 5 minutes = 300 s. If V = 12 V, I = 4 A, and t = 300 s, then E = 12 × 4 × 300 = 14 400 J.
⚠️ Common Misconceptions
Watch Out!
Students often think that materials with a high specific heat capacity heat up quickly. Wrong: A high SHC means the material heats up quickly because it absorbs a lot of energy. Correct: A high SHC means the material needs a lot of energy to change its temperature, so it heats up slowly and cools down slowly. Water (SHC = 4200 J/kg°C) heats up much more slowly than copper (SHC = 385 J/kg°C) for the same energy input.
Students often think that specific heat capacity depends on the mass of the object. Wrong: A larger block of aluminium has a higher specific heat capacity than a smaller block because there is more mass. Correct: Specific heat capacity is a property of the material itself, not the object. It is defined as the energy needed per kilogram per degree Celsius. A 1 kg block and a 10 kg block of the same material have the same SHC. The total energy needed depends on mass, but the SHC does not change.
✍️ 6-Mark Question
Extended Answer Question
6 marks: Describe how you would investigate the specific heat capacity of a metal block and explain how to improve the accuracy of your result.
First, measure the mass of the metal block using an electronic balance. Insert an electric immersion heater and a thermometer into the pre-drilled holes in the block. Wrap the block tightly in insulating material such as bubble wrap or foam to reduce thermal energy loss to the surroundings. Record the initial temperature of the block using the thermometer. Connect the heater to a DC power supply with an ammeter in series and a voltmeter in parallel across the heater. Switch on the power supply and start a stopwatch at the same time. Record the ammeter reading (current) and voltmeter reading (voltage). After a set time such as 5 minutes (300 seconds), switch off the power supply and record the final temperature. Calculate the energy supplied to the heater using E = VIt. Calculate the SHC using c = E / (mΔT).
To improve accuracy, several steps can be taken. Use thicker or better-quality insulation around the block to reduce energy loss to the surroundings, which is the main source of error. Ensure the immersion heater and thermometer fit tightly into their holes so that thermal contact is good and less energy is conducted away from the heater before reaching the block. Use a lid on top of the block if it has a cavity, to reduce convection losses. Replace the standard thermometer with a temperature sensor and data logger, which provides more precise and continuous readings. Allow the block to cool between repeats and take an average of several results to reduce the effect of random errors.
Mark scheme: 2 marks for describing the method (mass, heater, insulation, E = VIt, c = E/mΔT), 2 marks for explaining sources of error (energy loss to surroundings), 2 marks for specific improvements (better insulation, tight fittings, data logger, repeats)
📊 AO3: Analyse & Evaluate
Analysis and Evaluation
A student investigated the SHC of an aluminium block and obtained the following results from three trials:
| Trial | Energy Supplied (J) | Mass (kg) | ΔT (°C) | Calculated SHC (J/kg°C) |
|---|---|---|---|---|
| 1 | 10 000 | 1.0 | 11.1 | 901 |
| 2 | 10 000 | 1.0 | 12.5 | 800 |
| 3 | 10 000 | 1.0 | 9.8 | 1020 |
The accepted value for the SHC of aluminium is 900 J/kg°C. Identify the anomalous result, explain why it is anomalous, and suggest what might have caused it.
Answer: Trial 2 gives an SHC of 800 J/kg°C, which is significantly lower than the other two trials (901 and 1020) and much lower than the accepted value. This is the anomalous result. The calculated SHC is too low because ΔT is too high (12.5°C instead of the expected ~11.1°C). This could have been caused by the thermometer not being properly inserted into the block, so it was measuring the temperature of the air or heater rather than the block itself. Alternatively, the insulation may have been missing or inadequate in this trial, allowing more heat to reach the thermometer directly from the heater rather than through the block. It could also be that the student started the stopwatch late or read the temperature at the wrong time. Trial 3 gives a slightly high SHC, likely due to greater heat loss than Trial 1.
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